Operating SystemsAddress Translation in Segmentation

Segmentation Address Translation

LevelIntermediate

Duration90 mins

TopicAddress Translation in Segmentation

5 / 5

Translation Example

Seeing the Complete Picture

We've examined each component of segmented address translation in isolation: segment number extraction, offset extraction, bounds checking, and physical address formation. Now it's time to synthesize this knowledge by tracing complete translation examples from start to finish.

Worked examples are invaluable for cementing understanding. By following concrete numbers through each step, seeing exactly how bits are extracted, how tables are indexed, and how addresses are computed, the abstract mechanisms become tangible and intuitive. These examples will prepare you to trace translations yourself, debug memory issues, and understand hardware documentation.

What You Will Learn

By the end of this page, you will be able to trace any logical address through the complete translation pipeline, calculate physical addresses by hand, identify and diagnose translation failures, and understand how real systems report addressing information.

Example System Configuration

Before working through examples, let's establish a concrete system configuration that we'll use throughout this page. This represents a simplified but realistic segmented memory system.

system_configuration.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
=== Example System Configuration ===
 
Logical Address Format:
┌────────────────────────────────────┐
│       16-bit Logical Address       │
├──────────────┬─────────────────────┤
│ Segment (4b) │    Offset (12b)     │
│  bits 15-12  │    bits 11-0        │
└──────────────┴─────────────────────┘
 
Physical Address: 24 bits (16 MB addressable)
 
Key Parameters:
  - Maximum segments per process: 16 (2^4)
  - Maximum segment size: 4096 bytes (2^12)
  - Physical memory: 16 MB (2^24)
 
Segment Table for Process P (5 valid segments):
 
┌─────┬────────────┬───────┬─────────┬────────────┐
│ Seg │    Base    │ Limit │ Present │ Protection │
├─────┼────────────┼───────┼─────────┼────────────┤
│  0  │ 0x100000   │ 4096  │   Yes   │    R-X     │  Code
│  1  │ 0x200000   │ 8192  │   Yes   │    RW-     │  Data
│  2  │ 0x300000   │ 2048  │   Yes   │    RW-     │  Stack
│  3  │ 0x400000   │ 16384 │   Yes   │    RW-     │  Heap
│  4  │ 0x000000   │ 4096  │   No    │    RW-     │  Swapped
└─────┴────────────┴───────┴─────────┴────────────┘
 
STBR (Segment Table Base Register): 0x800000
STLR (Segment Table Length Register): 5
 
Notes:
  - Segment 0 (Code): Read & Execute, starts at 1 MB
  - Segment 1 (Data): Read & Write, starts at 2 MB  
  - Segment 2 (Stack): Read & Write, starts at 3 MB
  - Segment 3 (Heap): Read & Write, starts at 4 MB
  - Segment 4 (Extra): Currently swapped to disk (not present)

Quick Reference: Extraction Formulas
Component	Formula	Example (addr = 0x3A5F)
Segment Number	addr >> 12	0x3A5F >> 12 = 0x3
Offset	addr & 0xFFF	0x3A5F & 0xFFF = 0xA5F
Physical Address	base + offset	(from table) + 0xA5F

Why These Parameters?

The 4-bit segment / 12-bit offset split is chosen for pedagogical clarity. With only 16 possible segments and 4 KB maximum segment size, all values are small enough to trace comfortably. Real systems like Intel x86 use larger values but the same principles apply.

Example 1: Successful Translation Walkthrough

Let's trace a successful memory access that passes all checks and produces a valid physical address.

Scenario: A program executes MOV AX, [0x1800] — reading from logical address 0x1800.

Step-by-Step Translation:

example1_successful.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
=== Translation of Logical Address 0x1800 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x1800
Binary:          0001 1000 0000 0000
                 ├──┘ └──────────────┤
                   │          │
                   │          └── Offset bits (12 bits)
                   │
                   └── Segment bits (4 bits)
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0x1800 >> 12
               = 0x1800 / 4096
               = 0x1 (segment 1)
 
Calculation:
  0x1800 = 6144 decimal
  6144 / 4096 = 1 (integer division)
  
STEP 3: Extract Offset
─────────────────────────────────────────────
Offset = 0x1800 & 0x0FFF
       = 0x800
       = 2048 decimal
 
Calculation:
  0x1800 AND 0x0FFF:
    0001 1000 0000 0000
  & 0000 1111 1111 1111
  ─────────────────────
    0000 1000 0000 0000 = 0x800
 
STEP 4: Validate Segment Number
─────────────────────────────────────────────
Check: segment (1) < STLR (5)?
Answer: 1 < 5 ✓ VALID
 
The segment number is within the valid range.
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment Table Entry for Segment 1:
  Base:       0x200000 (2 MB)
  Limit:      8192 bytes
  Present:    Yes
  Protection: RW- (Read, Write, no Execute)
 
Entry address = STBR + (segment_num × entry_size)
              = 0x800000 + (1 × 8)
              = 0x800008
 
STEP 6: Check Present Bit
─────────────────────────────────────────────
Is segment present in memory?
Answer: Yes ✓ VALID
 
STEP 7: Check Bounds
─────────────────────────────────────────────
Check: offset (2048) < limit (8192)?
Answer: 2048 < 8192 ✓ VALID
 
The offset is within the segment's bounds.
 
STEP 8: Form Physical Address
─────────────────────────────────────────────
Physical Address = Base + Offset
                 = 0x200000 + 0x800
                 = 0x200800
 
Calculation:
    0x200000   (base: 2,097,152)
  + 0x000800   (offset: 2,048)
  ──────────
    0x200800   (physical: 2,099,200)
 
═══════════════════════════════════════════════
FINAL RESULT: 
  Logical  0x1800 → Physical 0x200800 ✓ SUCCESS
═══════════════════════════════════════════════

Converting Mermaid diagram...

Translation Complete

The logical address 0x1800 successfully translates to physical address 0x200800. The CPU can now access the memory at this physical location. This entire process happens in hardware within a few clock cycles.

Example 2: Bounds Violation

Now let's trace an address that fails bounds checking—a common programming error that results in a segmentation fault.

Scenario: A program has an array out-of-bounds bug and accesses logical address 0x2900 within the stack segment.

Step-by-Step Translation:

example2_bounds_fault.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
=== Translation of Logical Address 0x2900 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x2900
Binary:          0010 1001 0000 0000
                 ├──┘ └──────────────┤
                 Seg      Offset
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0x2900 >> 12
               = 0x2 (segment 2 - Stack)
 
STEP 3: Extract Offset  
─────────────────────────────────────────────
Offset = 0x2900 & 0x0FFF
       = 0x900
       = 2304 decimal
 
STEP 4: Validate Segment Number
─────────────────────────────────────────────
Check: segment (2) < STLR (5)?
Answer: 2 < 5 ✓ VALID
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment 2 (Stack):
  Base:       0x300000 (3 MB)
  Limit:      2048 bytes     ← Note: only 2048!
  Present:    Yes
  Protection: RW-
 
STEP 6: Check Present Bit  
─────────────────────────────────────────────
Is segment present? Yes ✓ VALID
 
STEP 7: Check Bounds ← FAILURE OCCURS HERE
─────────────────────────────────────────────
Check: offset (2304) < limit (2048)?
 
  Offset:    2304 (0x900)
  Limit:     2048 (0x800)
  
  2304 < 2048 ?
  Answer: NO ✗ FAULT!
 
The offset exceeds the segment's limit!
 
═══════════════════════════════════════════════
FAULT: SEGMENT BOUNDS VIOLATION
═══════════════════════════════════════════════
 
What happens next:
  1. MMU asserts fault signal
  2. CPU saves state (registers, flags, PC)
  3. CPU switches to kernel mode
  4. Control jumps to exception handler
  5. OS examines fault type and address
  
Possible OS responses:
  - If stack growth possible: extend stack, retry
  - If actual bug: deliver SIGSEGV to process
  - Default: process terminated, core dump generated
 
Error message might look like:
  "Segmentation fault (core dumped)"
  or
  "SIGSEGV at address 0x2900, stack segment overflow"

Analysis:

This failure represents a common bug: stack overflow. The stack segment was allocated only 2048 bytes (offsets 0x000 to 0x7FF), but the program tried to access offset 0x900 (2304). This could happen from:

Deep recursive function calls exceeding stack space
Large local arrays allocating more stack than available
Buffer overflow writing past local variables

The bounds check caught this before any memory corruption could occur.

Protection in Action

Without bounds checking, this access would have proceeded to physical address 0x300900, which lies in a different segment's memory (or unmapped space). This could corrupt data, leak sensitive information, or enable code injection. Bounds checking prevents all of these outcomes.

Example 3: Invalid Segment Number

Another failure case: referencing a segment number that doesn't exist in the segment table.

Scenario: A corrupted pointer contains the value 0xA100, referencing segment 10 which doesn't exist.

example3_invalid_segment.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
=== Translation of Logical Address 0xA100 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0xA100
Binary:          1010 0001 0000 0000
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0xA100 >> 12
               = 0xA (10 decimal)
 
STEP 3: Extract Offset
─────────────────────────────────────────────
Offset = 0xA100 & 0x0FFF
       = 0x100
       = 256 decimal
 
STEP 4: Validate Segment Number ← FAILURE HERE
─────────────────────────────────────────────
Check: segment (10) < STLR (5)?
 
  Segment Number: 10
  STLR (valid range): 5
  
  10 < 5 ?
  Answer: NO ✗ FAULT!
 
═══════════════════════════════════════════════
FAULT: INVALID SEGMENT NUMBER
═══════════════════════════════════════════════
 
The segment number exceeds the segment table length.
No segment table entry exists for segment 10.
 
This typically indicates:
  - Wild/corrupted pointer
  - Uninitialized pointer variable
  - Use-after-free bug
  - Stack smashing corrupted saved pointer
  - Deliberate attack attempt
 
OS Response:
  - Cannot resolve: no valid segment metadata
  - Deliver SIGSEGV to process
  - Process terminated

Segment Validity for Different Addresses
Address	Segment #	STLR Check	Result
0x0500	0	0 < 5 ✓	Valid segment
0x1A00	1	1 < 5 ✓	Valid segment
0x4FFF	4	4 < 5 ✓	Valid segment
0x5000	5	5 < 5 ✗	Invalid segment
0xFFFF	15	15 < 5 ✗	Invalid segment

Early Failure is Good

Notice that this fault occurs before any segment table access. The STLR check prevents reading invalid memory when looking up the segment descriptor. This is an important defense: even a corrupted segment table can't cause the translation hardware to access arbitrary memory.

Example 4: Segment Not Present

What happens when a valid segment number references a segment that has been swapped out to disk?

Scenario: The program accesses logical address 0x4200, referencing segment 4 which exists but is currently swapped out.

example4_not_present.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
=== Translation of Logical Address 0x4200 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x4200
Segment Number:  4
Offset:          0x200 (512 decimal)
 
STEP 2-4: Segment Number Valid
─────────────────────────────────────────────
Check: segment (4) < STLR (5)?
Answer: 4 < 5 ✓ VALID
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment 4:
  Base:       0x000000 (invalid - not loaded)
  Limit:      4096 bytes
  Present:    NO ← Segment is swapped out!
  Protection: RW-
 
STEP 6: Check Present Bit ← FAULT HERE
─────────────────────────────────────────────
Is segment present in memory?
Answer: NO ✗ FAULT!
 
═══════════════════════════════════════════════
FAULT: SEGMENT NOT PRESENT
═══════════════════════════════════════════════
 
This is a "soft" fault - potentially resolvable:
 
  1. CPU traps to kernel
  2. Kernel identifies fault type: missing segment
  3. Kernel locates segment on disk
  4. Kernel finds free physical memory
  5. Kernel loads segment from disk to memory
  6. Kernel updates segment table entry:
       - Base = new physical location
       - Present = Yes
  7. Kernel resumes the process
  8. Instruction is restarted
  9. Translation now succeeds!
 
This is the mechanism for demand segmentation (similar to
demand paging, but at segment granularity).
 
Timeline:
  [Fault] → [Disk I/O: ~10ms] → [Table Update] → [Resume] → [Success]

Key Difference from Previous Faults:

Unlike bounds violations or invalid segment numbers (which indicate bugs), a "segment not present" fault is a normal part of system operation when demand segmentation or swapping is used. The OS handles it transparently:

The process is blocked while I/O occurs
Another process runs during the wait
Once loaded, the original process resumes
From the program's perspective, the access eventually succeeds

This is similar to how demand paging works with page faults.

Demand Segmentation

Systems using demand segmentation only load segments when first accessed. A newly forked process might have all segments marked 'not present', loading them from the parent or disk only on demand. This reduces memory usage and startup time for large programs.

Quick Reference: Multiple Address Translations

Here's a quick reference showing multiple translations using our example segment table, demonstrating the variety of outcomes possible.

Complete Translation Examples
Logical	Seg	Offset	Base	Limit	Result
0x0100	0	0x100	0x100000	4096	Physical: 0x100100 ✓
0x0FFF	0	0xFFF	0x100000	4096	Physical: 0x100FFF ✓
0x1000	1	0x000	0x200000	8192	Physical: 0x200000 ✓
0x1800	1	0x800	0x200000	8192	Physical: 0x200800 ✓
0x1FFF	1	0xFFF	0x200000	8192	Physical: 0x200FFF ✓
0x2000	2	0x000	0x300000	2048	Physical: 0x300000 ✓
0x27FF	2	0x7FF	0x300000	2048	Physical: 0x3007FF ✓
0x2800	2	0x800		2048	FAULT: 0x800 ≥ 2048 ✗
0x3A5F	3	0xA5F	0x400000	16384	Physical: 0x400A5F ✓
0x4000	4				FAULT: Not Present ✗
0x5000	5				FAULT: Seg 5 ≥ STLR ✗
0xFFFF	15				FAULT: Seg 15 ≥ STLR ✗

batch_translation.c
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
#include <stdio.h>
#include <stdint.h>
 
/*
 * Batch Translation Example
 * Demonstrates multiple translations with various outcomes
 */
 
typedef struct {
    uint32_t base;
    uint32_t limit;
    int present;
} SegEntry;
 
SegEntry seg_table[] = {
    {0x100000, 4096,  1},  // Seg 0: Code
    {0x200000, 8192,  1},  // Seg 1: Data
    {0x300000, 2048,  1},  // Seg 2: Stack
    {0x400000, 16384, 1},  // Seg 3: Heap
    {0x000000, 4096,  0},  // Seg 4: Swapped
};
#define STLR 5
 
const char* translate(uint16_t logical, uint32_t* physical) {
    uint16_t seg = logical >> 12;
    uint16_t off = logical & 0x0FFF;
    
    if (seg >= STLR) return "Invalid segment";
    if (!seg_table[seg].present) return "Not present";
    if (off >= seg_table[seg].limit) return "Bounds violation";
    
    *physical = seg_table[seg].base + off;
    return "Success";
}
 
int main() {
    uint16_t test_addresses[] = {
        0x0100, 0x0FFF, 0x1000, 0x1800, 0x1FFF,
        0x2000, 0x27FF, 0x2800, 0x3A5F,
        0x4000, 0x5000, 0xFFFF
    };
    int num_tests = sizeof(test_addresses)/sizeof(test_addresses[0]);
    
    printf("%-10s %-4s %-6s %-10s %-10s %s
",
           "Logical", "Seg", "Offset", "Base", "Physical", "Status");
    printf("%-10s %-4s %-6s %-10s %-10s %s
",
           "-------", "---", "------", "----", "--------", "------");
    
    for (int i = 0; i < num_tests; i++) {
        uint16_t addr = test_addresses[i];
        uint32_t physical = 0;
        const char* status = translate(addr, &physical);
        
        uint16_t seg = addr >> 12;
        uint16_t off = addr & 0x0FFF;
        
        printf("0x%04X     %-4u 0x%03X  ", addr, seg, off);
        
        if (seg < STLR && seg_table[seg].present) {
            printf("0x%06X   ", seg_table[seg].base);
        } else {
            printf("%-10s ", "-");
        }
        
        if (physical != 0) {
            printf("0x%06X   ", physical);
        } else {
            printf("%-10s ", "-");
        }
        
        printf("%s
", status);
    }
    
    return 0;
}

Pattern Recognition

Notice how the segment number is always addr >> 12 (equivalent to the first hex digit). Mental math tip: for addresses like 0x3A5F, just look at the leading digit (3) to immediately know the segment. The remaining digits (A5F) are the offset.

Calculation Practice and Techniques

Being able to perform these calculations quickly is valuable for debugging, interview questions, and understanding system behavior. Here are techniques for fast mental calculation.

Quick Calculation Techniques

•Segment extraction: For 4-bit segment / 12-bit offset, the segment is simply the leftmost hex digit. 0x3A5F → segment 3.
•Offset extraction: The remaining hex digits are the offset. 0x3A5F → offset 0xA5F = 2655.
•Simple addition: To add offset to base, align columns. 0x400000 + 0xA5F = 0x400A5F (just concatenate when offset has fewer digits).
•Bounds check shortcut: Compare offset's high nibble to limit. If offset is 0x7xx and limit is 2048 (0x800), 0x7xx < 0x800 ✓.
•Hex to decimal for limits: 0x1000 = 4096, 0x2000 = 8192, 0x4000 = 16384. Know these powers of 2.

calculation_practice.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
=== Practice Problems ===
 
Given the segment table from our examples, translate these:
 
Problem 1: Logical Address 0x0ABC
───────────────────────────────
Your work:
  Segment = 0x0ABC >> 12 = ___
  Offset  = 0x0ABC & 0xFFF = ___
  Base    = _______ (look up seg ___)
  Limit   = _______ 
  Bounds check: ___ < ___ ?
  Physical = _______
 
Answer: Segment 0, Offset 0xABC, Base 0x100000
        Bounds: 0xABC (2748) < 4096 ✓
        Physical: 0x100000 + 0xABC = 0x100ABC
 
─────────────────────────────────────────────
 
Problem 2: Logical Address 0x3FFF
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Base    = _______
  Limit   = _______
  Bounds check: ___ < ___ ?
  Physical = _______
 
Answer: Segment 3, Offset 0xFFF (4095)
        Base 0x400000, Limit 16384
        Bounds: 4095 < 16384 ✓
        Physical: 0x400FFF
 
─────────────────────────────────────────────
 
Problem 3: Logical Address 0x1FFF (edge case)
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Limit check: ___ < ___ ?
 
Answer: Segment 1, Offset 0xFFF (4095)
        Limit 8192
        Bounds: 4095 < 8192 ✓
        Physical: 0x200FFF
        (This is valid - well within bounds)
 
─────────────────────────────────────────────
 
Problem 4: Logical Address 0x2900 (failure case)
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Limit = ___
  Bounds check: ___ < ___ ?
 
Answer: Segment 2, Offset 0x900 (2304)
        Limit 2048
        Bounds: 2304 < 2048 ✗ FAIL
        Result: Bounds violation fault

Exam Strategy

In exams, always show your work clearly: (1) Extract segment and offset, (2) State the base and limit from the table, (3) Show the bounds comparison, (4) Calculate the physical address. Even if you make an arithmetic error, showing the correct process often earns partial credit.

Summary: Mastering Translation Examples

Working through concrete examples transforms abstract understanding into practical skill. You can now trace any logical address through the complete segmented translation pipeline.

Key Takeaways

•Successful translation follows all steps: extract segment, extract offset, validate segment, check presence, check bounds, compute physical = base + offset
•Bounds violations occur when offset ≥ limit, indicating buffer overflows, stack overflow, or other bugs
•Invalid segment numbers (≥ STLR) indicate corrupted pointers or uninitialized variables
•Not-present faults are often resolvable by loading the segment from disk
•Quick mental calculation uses the hexadecimal structure: first digit(s) = segment, remaining = offset

Module Complete:

You have now completed the Address Translation module for Segmentation. You understand:

How segment numbers identify logical program units and index the segment table
How offsets specify byte positions within segments
How bounds checking protects memory integrity
How physical addresses are formed by adding base to offset
How to trace complete translations with concrete examples

This knowledge forms the foundation for understanding more advanced topics like segmentation with paging, the Intel x86 memory model, and the historical evolution from segmentation to modern virtual memory systems.

Module Complete

Congratulations! You've mastered segmented address translation. You can now extract address components, validate accesses, compute physical addresses, and diagnose translation failures. This understanding is fundamental to operating systems, systems programming, and computer architecture.

5 / 5

Loading learning content...

Operating SystemsAddress Translation in Segmentation

Segmentation Address Translation

LevelIntermediate

Duration90 mins

TopicAddress Translation in Segmentation

5 / 5

Translation Example

Seeing the Complete Picture

What You Will Learn

Example System Configuration

Before working through examples, let's establish a concrete system configuration that we'll use throughout this page. This represents a simplified but realistic segmented memory system.

system_configuration.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
=== Example System Configuration ===
 
Logical Address Format:
┌────────────────────────────────────┐
│       16-bit Logical Address       │
├──────────────┬─────────────────────┤
│ Segment (4b) │    Offset (12b)     │
│  bits 15-12  │    bits 11-0        │
└──────────────┴─────────────────────┘
 
Physical Address: 24 bits (16 MB addressable)
 
Key Parameters:
  - Maximum segments per process: 16 (2^4)
  - Maximum segment size: 4096 bytes (2^12)
  - Physical memory: 16 MB (2^24)
 
Segment Table for Process P (5 valid segments):
 
┌─────┬────────────┬───────┬─────────┬────────────┐
│ Seg │    Base    │ Limit │ Present │ Protection │
├─────┼────────────┼───────┼─────────┼────────────┤
│  0  │ 0x100000   │ 4096  │   Yes   │    R-X     │  Code
│  1  │ 0x200000   │ 8192  │   Yes   │    RW-     │  Data
│  2  │ 0x300000   │ 2048  │   Yes   │    RW-     │  Stack
│  3  │ 0x400000   │ 16384 │   Yes   │    RW-     │  Heap
│  4  │ 0x000000   │ 4096  │   No    │    RW-     │  Swapped
└─────┴────────────┴───────┴─────────┴────────────┘
 
STBR (Segment Table Base Register): 0x800000
STLR (Segment Table Length Register): 5
 
Notes:
  - Segment 0 (Code): Read & Execute, starts at 1 MB
  - Segment 1 (Data): Read & Write, starts at 2 MB  
  - Segment 2 (Stack): Read & Write, starts at 3 MB
  - Segment 3 (Heap): Read & Write, starts at 4 MB
  - Segment 4 (Extra): Currently swapped to disk (not present)

Quick Reference: Extraction Formulas
Component	Formula	Example (addr = 0x3A5F)
Segment Number	addr >> 12	0x3A5F >> 12 = 0x3
Offset	addr & 0xFFF	0x3A5F & 0xFFF = 0xA5F
Physical Address	base + offset	(from table) + 0xA5F

Why These Parameters?

Example 1: Successful Translation Walkthrough

Let's trace a successful memory access that passes all checks and produces a valid physical address.

Scenario: A program executes MOV AX, [0x1800] — reading from logical address 0x1800.

Step-by-Step Translation:

example1_successful.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
=== Translation of Logical Address 0x1800 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x1800
Binary:          0001 1000 0000 0000
                 ├──┘ └──────────────┤
                   │          │
                   │          └── Offset bits (12 bits)
                   │
                   └── Segment bits (4 bits)
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0x1800 >> 12
               = 0x1800 / 4096
               = 0x1 (segment 1)
 
Calculation:
  0x1800 = 6144 decimal
  6144 / 4096 = 1 (integer division)
  
STEP 3: Extract Offset
─────────────────────────────────────────────
Offset = 0x1800 & 0x0FFF
       = 0x800
       = 2048 decimal
 
Calculation:
  0x1800 AND 0x0FFF:
    0001 1000 0000 0000
  & 0000 1111 1111 1111
  ─────────────────────
    0000 1000 0000 0000 = 0x800
 
STEP 4: Validate Segment Number
─────────────────────────────────────────────
Check: segment (1) < STLR (5)?
Answer: 1 < 5 ✓ VALID
 
The segment number is within the valid range.
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment Table Entry for Segment 1:
  Base:       0x200000 (2 MB)
  Limit:      8192 bytes
  Present:    Yes
  Protection: RW- (Read, Write, no Execute)
 
Entry address = STBR + (segment_num × entry_size)
              = 0x800000 + (1 × 8)
              = 0x800008
 
STEP 6: Check Present Bit
─────────────────────────────────────────────
Is segment present in memory?
Answer: Yes ✓ VALID
 
STEP 7: Check Bounds
─────────────────────────────────────────────
Check: offset (2048) < limit (8192)?
Answer: 2048 < 8192 ✓ VALID
 
The offset is within the segment's bounds.
 
STEP 8: Form Physical Address
─────────────────────────────────────────────
Physical Address = Base + Offset
                 = 0x200000 + 0x800
                 = 0x200800
 
Calculation:
    0x200000   (base: 2,097,152)
  + 0x000800   (offset: 2,048)
  ──────────
    0x200800   (physical: 2,099,200)
 
═══════════════════════════════════════════════
FINAL RESULT: 
  Logical  0x1800 → Physical 0x200800 ✓ SUCCESS
═══════════════════════════════════════════════

Converting Mermaid diagram...

Translation Complete

Example 2: Bounds Violation

Now let's trace an address that fails bounds checking—a common programming error that results in a segmentation fault.

Scenario: A program has an array out-of-bounds bug and accesses logical address 0x2900 within the stack segment.

Step-by-Step Translation:

example2_bounds_fault.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
=== Translation of Logical Address 0x2900 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x2900
Binary:          0010 1001 0000 0000
                 ├──┘ └──────────────┤
                 Seg      Offset
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0x2900 >> 12
               = 0x2 (segment 2 - Stack)
 
STEP 3: Extract Offset  
─────────────────────────────────────────────
Offset = 0x2900 & 0x0FFF
       = 0x900
       = 2304 decimal
 
STEP 4: Validate Segment Number
─────────────────────────────────────────────
Check: segment (2) < STLR (5)?
Answer: 2 < 5 ✓ VALID
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment 2 (Stack):
  Base:       0x300000 (3 MB)
  Limit:      2048 bytes     ← Note: only 2048!
  Present:    Yes
  Protection: RW-
 
STEP 6: Check Present Bit  
─────────────────────────────────────────────
Is segment present? Yes ✓ VALID
 
STEP 7: Check Bounds ← FAILURE OCCURS HERE
─────────────────────────────────────────────
Check: offset (2304) < limit (2048)?
 
  Offset:    2304 (0x900)
  Limit:     2048 (0x800)
  
  2304 < 2048 ?
  Answer: NO ✗ FAULT!
 
The offset exceeds the segment's limit!
 
═══════════════════════════════════════════════
FAULT: SEGMENT BOUNDS VIOLATION
═══════════════════════════════════════════════
 
What happens next:
  1. MMU asserts fault signal
  2. CPU saves state (registers, flags, PC)
  3. CPU switches to kernel mode
  4. Control jumps to exception handler
  5. OS examines fault type and address
  
Possible OS responses:
  - If stack growth possible: extend stack, retry
  - If actual bug: deliver SIGSEGV to process
  - Default: process terminated, core dump generated
 
Error message might look like:
  "Segmentation fault (core dumped)"
  or
  "SIGSEGV at address 0x2900, stack segment overflow"

Analysis:

Deep recursive function calls exceeding stack space
Large local arrays allocating more stack than available
Buffer overflow writing past local variables

The bounds check caught this before any memory corruption could occur.

Protection in Action

Example 3: Invalid Segment Number

Another failure case: referencing a segment number that doesn't exist in the segment table.

Scenario: A corrupted pointer contains the value 0xA100, referencing segment 10 which doesn't exist.

example3_invalid_segment.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
=== Translation of Logical Address 0xA100 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0xA100
Binary:          1010 0001 0000 0000
 
STEP 2: Extract Segment Number
─────────────────────────────────────────────
Segment Number = 0xA100 >> 12
               = 0xA (10 decimal)
 
STEP 3: Extract Offset
─────────────────────────────────────────────
Offset = 0xA100 & 0x0FFF
       = 0x100
       = 256 decimal
 
STEP 4: Validate Segment Number ← FAILURE HERE
─────────────────────────────────────────────
Check: segment (10) < STLR (5)?
 
  Segment Number: 10
  STLR (valid range): 5
  
  10 < 5 ?
  Answer: NO ✗ FAULT!
 
═══════════════════════════════════════════════
FAULT: INVALID SEGMENT NUMBER
═══════════════════════════════════════════════
 
The segment number exceeds the segment table length.
No segment table entry exists for segment 10.
 
This typically indicates:
  - Wild/corrupted pointer
  - Uninitialized pointer variable
  - Use-after-free bug
  - Stack smashing corrupted saved pointer
  - Deliberate attack attempt
 
OS Response:
  - Cannot resolve: no valid segment metadata
  - Deliver SIGSEGV to process
  - Process terminated

Segment Validity for Different Addresses
Address	Segment #	STLR Check	Result
0x0500	0	0 < 5 ✓	Valid segment
0x1A00	1	1 < 5 ✓	Valid segment
0x4FFF	4	4 < 5 ✓	Valid segment
0x5000	5	5 < 5 ✗	Invalid segment
0xFFFF	15	15 < 5 ✗	Invalid segment

Early Failure is Good

Example 4: Segment Not Present

What happens when a valid segment number references a segment that has been swapped out to disk?

Scenario: The program accesses logical address 0x4200, referencing segment 4 which exists but is currently swapped out.

example4_not_present.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
=== Translation of Logical Address 0x4200 ===
 
STEP 1: Parse the Logical Address
─────────────────────────────────────────────
Logical Address: 0x4200
Segment Number:  4
Offset:          0x200 (512 decimal)
 
STEP 2-4: Segment Number Valid
─────────────────────────────────────────────
Check: segment (4) < STLR (5)?
Answer: 4 < 5 ✓ VALID
 
STEP 5: Look Up Segment Table Entry
─────────────────────────────────────────────
Segment 4:
  Base:       0x000000 (invalid - not loaded)
  Limit:      4096 bytes
  Present:    NO ← Segment is swapped out!
  Protection: RW-
 
STEP 6: Check Present Bit ← FAULT HERE
─────────────────────────────────────────────
Is segment present in memory?
Answer: NO ✗ FAULT!
 
═══════════════════════════════════════════════
FAULT: SEGMENT NOT PRESENT
═══════════════════════════════════════════════
 
This is a "soft" fault - potentially resolvable:
 
  1. CPU traps to kernel
  2. Kernel identifies fault type: missing segment
  3. Kernel locates segment on disk
  4. Kernel finds free physical memory
  5. Kernel loads segment from disk to memory
  6. Kernel updates segment table entry:
       - Base = new physical location
       - Present = Yes
  7. Kernel resumes the process
  8. Instruction is restarted
  9. Translation now succeeds!
 
This is the mechanism for demand segmentation (similar to
demand paging, but at segment granularity).
 
Timeline:
  [Fault] → [Disk I/O: ~10ms] → [Table Update] → [Resume] → [Success]

Key Difference from Previous Faults:

The process is blocked while I/O occurs
Another process runs during the wait
Once loaded, the original process resumes
From the program's perspective, the access eventually succeeds

This is similar to how demand paging works with page faults.

Demand Segmentation

Quick Reference: Multiple Address Translations

Here's a quick reference showing multiple translations using our example segment table, demonstrating the variety of outcomes possible.

Complete Translation Examples
Logical	Seg	Offset	Base	Limit	Result
0x0100	0	0x100	0x100000	4096	Physical: 0x100100 ✓
0x0FFF	0	0xFFF	0x100000	4096	Physical: 0x100FFF ✓
0x1000	1	0x000	0x200000	8192	Physical: 0x200000 ✓
0x1800	1	0x800	0x200000	8192	Physical: 0x200800 ✓
0x1FFF	1	0xFFF	0x200000	8192	Physical: 0x200FFF ✓
0x2000	2	0x000	0x300000	2048	Physical: 0x300000 ✓
0x27FF	2	0x7FF	0x300000	2048	Physical: 0x3007FF ✓
0x2800	2	0x800		2048	FAULT: 0x800 ≥ 2048 ✗
0x3A5F	3	0xA5F	0x400000	16384	Physical: 0x400A5F ✓
0x4000	4				FAULT: Not Present ✗
0x5000	5				FAULT: Seg 5 ≥ STLR ✗
0xFFFF	15				FAULT: Seg 15 ≥ STLR ✗

batch_translation.c
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
#include <stdio.h>
#include <stdint.h>
 
/*
 * Batch Translation Example
 * Demonstrates multiple translations with various outcomes
 */
 
typedef struct {
    uint32_t base;
    uint32_t limit;
    int present;
} SegEntry;
 
SegEntry seg_table[] = {
    {0x100000, 4096,  1},  // Seg 0: Code
    {0x200000, 8192,  1},  // Seg 1: Data
    {0x300000, 2048,  1},  // Seg 2: Stack
    {0x400000, 16384, 1},  // Seg 3: Heap
    {0x000000, 4096,  0},  // Seg 4: Swapped
};
#define STLR 5
 
const char* translate(uint16_t logical, uint32_t* physical) {
    uint16_t seg = logical >> 12;
    uint16_t off = logical & 0x0FFF;
    
    if (seg >= STLR) return "Invalid segment";
    if (!seg_table[seg].present) return "Not present";
    if (off >= seg_table[seg].limit) return "Bounds violation";
    
    *physical = seg_table[seg].base + off;
    return "Success";
}
 
int main() {
    uint16_t test_addresses[] = {
        0x0100, 0x0FFF, 0x1000, 0x1800, 0x1FFF,
        0x2000, 0x27FF, 0x2800, 0x3A5F,
        0x4000, 0x5000, 0xFFFF
    };
    int num_tests = sizeof(test_addresses)/sizeof(test_addresses[0]);
    
    printf("%-10s %-4s %-6s %-10s %-10s %s
",
           "Logical", "Seg", "Offset", "Base", "Physical", "Status");
    printf("%-10s %-4s %-6s %-10s %-10s %s
",
           "-------", "---", "------", "----", "--------", "------");
    
    for (int i = 0; i < num_tests; i++) {
        uint16_t addr = test_addresses[i];
        uint32_t physical = 0;
        const char* status = translate(addr, &physical);
        
        uint16_t seg = addr >> 12;
        uint16_t off = addr & 0x0FFF;
        
        printf("0x%04X     %-4u 0x%03X  ", addr, seg, off);
        
        if (seg < STLR && seg_table[seg].present) {
            printf("0x%06X   ", seg_table[seg].base);
        } else {
            printf("%-10s ", "-");
        }
        
        if (physical != 0) {
            printf("0x%06X   ", physical);
        } else {
            printf("%-10s ", "-");
        }
        
        printf("%s
", status);
    }
    
    return 0;
}

Pattern Recognition

Calculation Practice and Techniques

Being able to perform these calculations quickly is valuable for debugging, interview questions, and understanding system behavior. Here are techniques for fast mental calculation.

Quick Calculation Techniques

•Segment extraction: For 4-bit segment / 12-bit offset, the segment is simply the leftmost hex digit. 0x3A5F → segment 3.
•Offset extraction: The remaining hex digits are the offset. 0x3A5F → offset 0xA5F = 2655.
•Simple addition: To add offset to base, align columns. 0x400000 + 0xA5F = 0x400A5F (just concatenate when offset has fewer digits).
•Bounds check shortcut: Compare offset's high nibble to limit. If offset is 0x7xx and limit is 2048 (0x800), 0x7xx < 0x800 ✓.
•Hex to decimal for limits: 0x1000 = 4096, 0x2000 = 8192, 0x4000 = 16384. Know these powers of 2.

calculation_practice.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
=== Practice Problems ===
 
Given the segment table from our examples, translate these:
 
Problem 1: Logical Address 0x0ABC
───────────────────────────────
Your work:
  Segment = 0x0ABC >> 12 = ___
  Offset  = 0x0ABC & 0xFFF = ___
  Base    = _______ (look up seg ___)
  Limit   = _______ 
  Bounds check: ___ < ___ ?
  Physical = _______
 
Answer: Segment 0, Offset 0xABC, Base 0x100000
        Bounds: 0xABC (2748) < 4096 ✓
        Physical: 0x100000 + 0xABC = 0x100ABC
 
─────────────────────────────────────────────
 
Problem 2: Logical Address 0x3FFF
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Base    = _______
  Limit   = _______
  Bounds check: ___ < ___ ?
  Physical = _______
 
Answer: Segment 3, Offset 0xFFF (4095)
        Base 0x400000, Limit 16384
        Bounds: 4095 < 16384 ✓
        Physical: 0x400FFF
 
─────────────────────────────────────────────
 
Problem 3: Logical Address 0x1FFF (edge case)
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Limit check: ___ < ___ ?
 
Answer: Segment 1, Offset 0xFFF (4095)
        Limit 8192
        Bounds: 4095 < 8192 ✓
        Physical: 0x200FFF
        (This is valid - well within bounds)
 
─────────────────────────────────────────────
 
Problem 4: Logical Address 0x2900 (failure case)
───────────────────────────────
Your work:
  Segment = ___
  Offset  = ___
  Limit = ___
  Bounds check: ___ < ___ ?
 
Answer: Segment 2, Offset 0x900 (2304)
        Limit 2048
        Bounds: 2304 < 2048 ✗ FAIL
        Result: Bounds violation fault

Exam Strategy

Summary: Mastering Translation Examples

Working through concrete examples transforms abstract understanding into practical skill. You can now trace any logical address through the complete segmented translation pipeline.

Key Takeaways

•Successful translation follows all steps: extract segment, extract offset, validate segment, check presence, check bounds, compute physical = base + offset
•Bounds violations occur when offset ≥ limit, indicating buffer overflows, stack overflow, or other bugs
•Invalid segment numbers (≥ STLR) indicate corrupted pointers or uninitialized variables
•Not-present faults are often resolvable by loading the segment from disk
•Quick mental calculation uses the hexadecimal structure: first digit(s) = segment, remaining = offset

Module Complete:

You have now completed the Address Translation module for Segmentation. You understand:

How segment numbers identify logical program units and index the segment table
How offsets specify byte positions within segments
How bounds checking protects memory integrity
How physical addresses are formed by adding base to offset
How to trace complete translations with concrete examples

Module Complete

5 / 5

Segmentation Address Translation - Page 5 of 5 | OneNoughtOne