Bit Shifting - Learning Module

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Shift Overflow and Loss of Bits

When Bits Vanish Into the Void

Bit shifting is elegant and fast—until you push it past its limits. Shift overflow and bit loss are the dark corners of this operation where bits silently disappear, values wrap unexpectedly, and undefined behavior lurks in wait for the unwary programmer.

Unlike arithmetic overflow (which often triggers exceptions or is detectable), shift-related data loss is typically silent. Your program continues running with corrupted data, producing wrong results that may not surface until much later—or ever, if you're unlucky enough. Understanding these edge cases is essential for writing robust bit manipulation code.

What You Will Learn

By the end of this page, you will understand exactly what happens when bits shift beyond the boundaries of their container, the difference between defined wraparound behavior and undefined behavior, how to detect when overflow has occurred or will occur, and how to write safe code that either prevents these issues or handles them gracefully.

Understanding Bit Loss in Left Shift

When you left-shift a value, bits move leftward and zeros fill from the right. But what happens to the bits that move beyond the leftmost position? They're discarded—gone forever, with no trace left behind.

In fixed-width integer types (like int32 or uint8), there's a finite container for bits. Left shifting eventually pushes significant bits out of this container:

Bit Loss in 8-bit Left Shift (value = 170 = 0xAA)
Shift	Binary Before	Binary After	Lost Bits	Decimal
<< 0	10101010	10101010	none	170
<< 1	10101010	01010100	1	84
<< 2	10101010	10101000	01	168
<< 3	10101010	01010000	101	80
<< 4	10101010	10100000	0101	160
<< 5	10101010	01000000	10101	64
<< 6	10101010	10000000	010101	128
<< 7	10101010	00000000	1010101	0
<< 8	10101010	00000000	10101010	0

Observe the progression: with each shift, more original bits are pushed out. By shift 7, only the original LSB remains (now in the MSB position). By shift 8, all original bits are gone—replaced entirely by zeros.

The mathematical perspective:

Left shift by k is equivalent to multiplication by 2^k, but with modular arithmetic:

(x << k) mod 2^n where n is the bit width

This modular behavior is why 170 << 1 = 340 mod 256 = 84 for an 8-bit type. The multiplication produces 340, but only the low 8 bits (340 & 0xFF = 84) are kept.

No Error, No Warning

In most languages, bit loss during left shift produces no runtime error. The operation 'succeeds' and returns a value—it's just not the value you might expect from mathematical multiplication. This silent failure makes shift overflow bugs particularly insidious.

left_shift_overflow.py
Python
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import numpy as np
 
def demonstrate_left_shift_overflow():
    """Show how left shift loses bits in fixed-width integers."""
    
    # Python's native int has arbitrary precision - no overflow
    x_python = 170
    print("Python (arbitrary precision):")
    for shift in range(10):
        result = x_python << shift
        print(f"  170 << {shift} = {result}")
    
    # NumPy integers have fixed width - they overflow/wrap
    print("\nNumPy uint8 (8-bit unsigned):")
    x_np = np.uint8(170)
    for shift in range(10):
        result = x_np << shift
        print(f"  170 << {shift} = {result} (binary: {format(result, '08b')})")
    
    print("\nNotice: Mathematical 170*2 = 340, but uint8 gives 84 (340 mod 256)")
 
def detect_left_shift_overflow_32bit(x: int, shift: int) -> tuple:
    """
    Detect if left shifting x by shift positions would overflow a 32-bit unsigned.
    
    Returns: (would_overflow, result_if_masked)
    """
    MAX_UINT32 = 0xFFFFFFFF
    
    if shift >= 32:
        # Shifting by 32+ always loses all bits
        return (x != 0, 0)
    
    # Check if any bits would be shifted out
    # Bits at positions (32 - shift) and above will be lost
    high_bits_mask = MAX_UINT32 << (32 - shift) if shift > 0 else 0
    
    would_overflow = (x & high_bits_mask) != 0
    result = (x << shift) & MAX_UINT32
    
    return (would_overflow, result)
 
if __name__ == "__main__":
    demonstrate_left_shift_overflow()
    
    print("\nOverflow detection examples:")
    test_cases = [
        (1, 30),   # 1 << 30 = 1073741824, fits in 32 bits
        (1, 31),   # 1 << 31 = 2147483648, fits in uint32
        (1, 32),   # Would need 33 bits
        (3, 31),   # 3 << 31 = 6442450944, needs 33 bits
        (0xFF, 25), # 0xFF << 25 = 8556380160, needs 34 bits
    ]
    
    for x, shift in test_cases:
        overflow, result = detect_left_shift_overflow_32bit(x, shift)
        status = "OVERFLOW" if overflow else "OK"
        print(f"  {x} << {shift}: {status}, masked result = {result}")

Understanding Bit Loss in Right Shift

Right shift loses bits from the opposite end—the low-order bits are pushed out and discarded. This is equivalent to integer division where the remainder is thrown away.

Despite the name 'overflow' being less commonly used for right shift, the concept of information loss is just as real. Once bits are shifted out, there's no way to recover the original value without external information.

Bit Loss in 8-bit Right Shift (value = 170 = 0xAA)
Shift	Binary Before	Binary After	Lost Bits	Decimal
0	10101010	10101010	none	170
1	10101010	01010101	0	85
2	10101010	00101010	10	42
3	10101010	00010101	010	21
4	10101010	00001010	1010	10
5	10101010	00000101	01010	5
6	10101010	00000010	101010	2
7	10101010	00000001	0101010	1
8	10101010	00000000	10101010	0

Key observations:

Truncation toward zero — Each right shift by 1 approximately halves the value, discarding any fractional part. 170 → 85 → 42 → 21 → 10 → 5 → 2 → 1 → 0.
Non-reversible — You cannot recover 170 from 85 using left shift. 85 << 1 = 170 works, but 21 << 3 = 168 ≠ 170. The lost bits are truly lost.
Total loss — Shifting by the bit width or more produces 0 for all positive values (or -1 for negative values with arithmetic shift).

The remainder connection:

The bits shifted out represent the remainder of division by 2^k. This can be captured using a mask:

x >> k gives the quotient of x ÷ 2^k
x & ((1 << k) - 1) gives the remainder of x ÷ 2^k

right_shift_information_loss.py
Python
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def demonstrate_information_loss():
    """Show that right shift is not reversible."""
    
    original = 170
    
    print("Right shift and attempted reversal:")
    print(f"Original: {original} = {bin(original)}")
    print()
    
    for shift in range(1, 8):
        shifted = original >> shift
        reversed_attempt = shifted << shift
        lost_bits = original & ((1 << shift) - 1)
        
        recoverable = "Yes" if reversed_attempt == original else "No"
        
        print(f"  >> {shift}: {shifted:3d} ({bin(shifted):>10s})")
        print(f"      << {shift}: {reversed_attempt:3d} ({bin(reversed_attempt):>10s}) - Matches original? {recoverable}")
        print(f"      Lost bits: {lost_bits} ({bin(lost_bits)})")
        print()
 
def extract_quotient_and_remainder(x: int, k: int) -> tuple:
    """
    Compute x ÷ 2^k and x mod 2^k using only bit operations.
    
    This demonstrates that right shift captures the quotient
    and masking captures the remainder.
    """
    quotient = x >> k
    remainder = x & ((1 << k) - 1)
    
    # Verify: x == quotient * 2^k + remainder
    assert x == (quotient << k) + remainder
    
    return (quotient, remainder)
 
if __name__ == "__main__":
    demonstrate_information_loss()
    
    print("Quotient and remainder extraction:")
    for x in [170, 255, 100, 7]:
        for k in [1, 2, 3, 4]:
            q, r = extract_quotient_and_remainder(x, k)
            divisor = 1 << k
            print(f"  {x} ÷ {divisor} = {q} remainder {r} (verify: {q}*{divisor}+{r} = {q*divisor+r})")
        print()

Capture Before You Lose

If you need the bits that would be shifted out (the remainder), extract them with a mask BEFORE performing the shift. Once the shift executes, those bits are gone. This is a common pattern in division algorithms and when converting bit positions.

Shift Amount Edge Cases

Beyond bit loss within valid shift amounts, there are dangerous edge cases when the shift amount itself is problematic. These cause the most serious bugs because the behavior varies dramatically across languages and platforms.

Dangerous Shift Amounts

•Shift by 0 — Well-defined everywhere: x << 0 = x and x >> 0 = x. Harmless but occasionally wasteful.
•Shift by bit width — x << 32 for a 32-bit integer. Undefined in C/C++! In Java/JavaScript, wraps to x << 0 = x.
•Shift by more than bit width — x << 40 for a 32-bit type. Also undefined in C/C++; wraps modulo 32/64 in Java/JS.
•Shift by negative amount — x << -1. Undefined in C/C++; in Java, -1 & 31 = 31, so it becomes x << 31.
•Shift with volatile or unknown amounts — When the shift amount is runtime-computed, all these edge cases become possible failure modes.

Shift Amount Behavior Across Languages (for 32-bit integers)
Expression	C/C++	Java	JavaScript (\|0)	Python
x << 0	x	x	x	x
x << 32	Undefined!	x << 0 = x	x << 0 = x	x × 2^32
x << 33	Undefined!	x << 1	x << 1	x × 2^33
x << 64	Undefined!	x << 0 = x	x << 0 = x	x × 2^64
x << -1	Undefined!	x << 31	x << 31	Error
x >> 32	Undefined!	x >> 0 = x	x >> 0 = x	0 or -1*

*Python: For arbitrary precision integers, right shift by any amount toward 0 (positive) or -1 (negative).

Why C/C++ is undefined:

The C standard leaves shift by >= width undefined to allow for efficient compilation on different hardware. Some CPUs (x86) mask the shift amount; others (some ARM modes) saturate it; others do something else entirely. By leaving it undefined, the compiler can generate the fastest code for the target architecture without guaranteeing specific behavior.

Why Java/JavaScript wrap:

These languages chose to define specific behavior for portability. The shift amount is masked with & 31 (for 32-bit) or & 63 (for 64-bit), effectively using only the low bits of the shift amount. This is predictable but can hide bugs if you expect x << 32 = 0.

shift_amount_safety.c
C
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#include <stdint.h>
#include <stdio.h>
#include <stdbool.h>
 
#define BITS_32 32
#define BITS_64 64
 
// UNSAFE: This function has undefined behavior for n >= 32
uint32_t unsafe_left_shift(uint32_t x, unsigned int n) {
    return x << n;  // UB if n >= 32
}
 
// SAFE: Explicitly handle edge cases
uint32_t safe_left_shift_32(uint32_t x, unsigned int n) {
    if (n >= BITS_32) {
        return 0;  // All bits shifted out
    }
    return x << n;
}
 
// SAFE: Match Java/JavaScript behavior (wrapping)
uint32_t java_style_left_shift(uint32_t x, int n) {
    // Java masks with & 31
    unsigned int effective_shift = n & 31;
    return x << effective_shift;
}
 
// SAFE: Saturating shift (clamp to valid range)
uint32_t saturating_left_shift(uint32_t x, unsigned int n) {
    if (n >= BITS_32) {
        n = BITS_32 - 1;  // Clamp to maximum valid shift
    }
    return x << n;
}
 
// SAFE: Check for potential overflow before shifting
bool would_left_shift_overflow(uint32_t x, unsigned int n, uint32_t *result) {
    if (n >= BITS_32) {
        *result = 0;
        return x != 0;  // Overflow if any bits were set
    }
    
    // Check if any of the bits that will be shifted out are set
    uint32_t bits_to_shift_out = x >> (BITS_32 - n);
    *result = x << n;
    
    return bits_to_shift_out != 0;
}
 
// SAFE: Right shift with bounds checking
uint32_t safe_right_shift_32(uint32_t x, unsigned int n) {
    if (n >= BITS_32) {
        return 0;  // All bits shifted out
    }
    return x >> n;
}
 
// SAFE: Signed arithmetic right shift with bounds checking
int32_t safe_arithmetic_right_shift(int32_t x, unsigned int n) {
    if (n >= BITS_32) {
        // All bits become the sign bit
        return x < 0 ? -1 : 0;
    }
    return x >> n;  // Implementation-defined but usually arithmetic
}
 
int main(void) {
    uint32_t x = 0x80000000;  // 2^31
    
    printf("Safe shift demonstrations:\n\n");
    
    printf("safe_left_shift_32(1, 31) = %u\n", safe_left_shift_32(1, 31));
    printf("safe_left_shift_32(1, 32) = %u\n", safe_left_shift_32(1, 32));
    printf("safe_left_shift_32(1, 100) = %u\n", safe_left_shift_32(1, 100));
    
    printf("\njava_style_left_shift(1, 32) = %u (wraps to << 0)\n", 
           java_style_left_shift(1, 32));
    printf("java_style_left_shift(1, 33) = %u (wraps to << 1)\n", 
           java_style_left_shift(1, 33));
    
    printf("\nOverflow detection:\n");
    uint32_t result;
    bool overflow = would_left_shift_overflow(x, 1, &result);
    printf("0x%08X << 1: overflow = %s, result = 0x%08X\n", 
           x, overflow ? "YES" : "NO", result);
    
    overflow = would_left_shift_overflow(0x7FFFFFFF, 1, &result);
    printf("0x7FFFFFFF << 1: overflow = %s, result = 0x%08X\n",
           overflow ? "YES" : "NO", result);
    
    return 0;
}

Never Trust User Input for Shift Amounts

If a shift amount comes from external input (user data, file, network), it could be any value—including negative numbers or values larger than the bit width. Always validate and clamp shift amounts before use. Failure to do so is a common source of security vulnerabilities in low-level code.

Signed Integer Overflow in Left Shift

The interaction between left shift and signed integers is particularly treacherous. In C/C++ (before C++20), left-shifting a positive signed value such that it 'overflows' into the sign bit is undefined behavior.

The problematic scenario:

int x = 1;         // A positive signed 32-bit integer
int y = x << 31;   // Shifts 1 into the sign bit position

Mathematically, 1 × 2^31 = 2147483648, which exceeds the maximum signed 32-bit value (2147483647). The result would be interpreted as a negative number in two's complement. In C/C++, this is undefined behavior.

Signed Left Shift Danger Zones (32-bit signed)
Expression	Mathematical Result	Max Signed 32-bit	Status
1 << 30	1,073,741,824	2,147,483,647	OK
1 << 31	2,147,483,648	2,147,483,647	Undefined!
2 << 30	2,147,483,648	2,147,483,647	Undefined!
3 << 30	3,221,225,472	2,147,483,647	Undefined!
0x40000000 << 1	2,147,483,648	2,147,483,647	Undefined!

Why this is undefined in C:

Signed integer overflow is undefined behavior in C because different hardware handles it differently (some wrap, some trap, some saturate). The standard doesn't impose a model to allow maximum performance.

What typically happens (but is not guaranteed):

On most systems with two's complement arithmetic and at low optimization levels, 1 << 31 produces -2147483648 (INT_MIN). But optimizing compilers may:

Assume the overflow never happens and optimize based on that
Produce different results at different optimization levels
Generate code that behaves differently than the source suggests

The C++20 fix:

C++20 finally defined the behavior: left-shifting a signed value produces the correctly wrapped two's complement result, as if the computation were done in unsigned and converted back. This matches what most hardware does anyway.

signed_shift_safety.c
C
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#include <stdint.h>
#include <limits.h>
#include <stdbool.h>
 
// Check if left-shifting a signed value would overflow
// Returns true if overflow would occur
bool would_signed_left_shift_overflow(int32_t x, unsigned int n) {
    if (n == 0) return false;
    if (n >= 32) return x != 0;  // Any shift by 32+ loses all bits
    
    // For negative values, left shift is always UB in pre-C++20
    if (x < 0) return true;
    
    // For positive values, check if result would exceed INT32_MAX
    // Maximum value that can be shifted by n without overflow:
    // INT32_MAX >> n
    int32_t max_shiftable = INT32_MAX >> n;
    
    return x > max_shiftable;
}
 
// Safe signed left shift - uses unsigned to avoid UB
// Returns the two's complement result
int32_t safe_signed_left_shift(int32_t x, unsigned int n) {
    if (n >= 32) return 0;
    
    // Convert to unsigned, shift, convert back
    // This is well-defined (though implementation-defined on conversion back)
    uint32_t ux = (uint32_t)x;
    uint32_t result = ux << n;
    return (int32_t)result;
}
 
// Checked signed left shift - returns success/failure
bool checked_signed_left_shift(int32_t x, unsigned int n, int32_t *result) {
    if (would_signed_left_shift_overflow(x, n)) {
        *result = 0;  // Or any sentinel value
        return false;  // Overflow would occur
    }
    
    *result = x << n;  // Safe: no overflow
    return true;  // Success
}
 
// Example usage
#include <stdio.h>
 
int main(void) {
    printf("Signed left shift overflow detection:\n\n");
    
    struct { int32_t x; unsigned int n; } tests[] = {
        {1, 30},          // OK: 1 << 30 = 1073741824
        {1, 31},          // Overflow: 1 << 31 = 2147483648 > INT_MAX
        {2, 30},          // Overflow: 2 << 30 = 2147483648 > INT_MAX
        {0x20000000, 1},  // OK: 0x20000000 << 1 = 0x40000000
        {0x40000000, 1},  // Overflow: 0x40000000 << 1 = 0x80000000 > INT_MAX
        {-1, 1},          // Undefined: negative left shift
    };
    
    for (size_t i = 0; i < sizeof(tests)/sizeof(tests[0]); i++) {
        int32_t x = tests[i].x;
        unsigned int n = tests[i].n;
        
        bool overflow = would_signed_left_shift_overflow(x, n);
        int32_t safe_result = safe_signed_left_shift(x, n);
        
        printf("  %d << %u: %s, safe result = %d\n",
               x, n, 
               overflow ? "OVERFLOW" : "OK",
               safe_result);
    }
    
    return 0;
}

The Safest Approach

For maximum safety and portability, always use unsigned types for bit manipulation. Cast signed values to unsigned before shifting, perform the operations, then cast back if needed. This sidesteps all signed overflow undefined behavior.

Detecting and Preventing Overflow

Rather than dealing with overflow after it happens (which may be too late), we can detect potential overflow before performing the shift. This is essential for robust code handling untrusted inputs.

Overflow Detection Strategies

•Pre-check the shift amount — If n >= bit_width, all significant bits will be lost. Handle this case explicitly.
•Pre-check high bits for left shift — If any of the top n bits are set before a left shift by n, those bits will be lost.
•Check result against expected mathematical result — For small values, you can compute using a wider type and compare.
•Use built-in overflow checks — Some compilers provide __builtin_mul_overflow and similar intrinsics.
•Use wider intermediate types — Compute in 64-bit, check if result fits in 32-bit, then narrow.

overflow_detection.py
Python
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from typing import Tuple
 
def safe_left_shift_32(x: int, n: int) -> Tuple[int, bool]:
    """
    Perform 32-bit unsigned left shift with overflow detection.
    
    Returns: (result, overflow_occurred)
    - result: The masked 32-bit result
    - overflow_occurred: True if bits were lost
    """
    if n < 0:
        raise ValueError("Negative shift amount")
    
    MAX_UINT32 = 0xFFFFFFFF
    
    # Clamp x to 32 bits first
    x = x & MAX_UINT32
    
    if n >= 32:
        # All bits shifted out
        return (0, x != 0)
    
    if n == 0:
        return (x, False)
    
    # Check if any bits in positions [32-n, 31] are set
    # These bits will be shifted out
    high_bits_mask = (MAX_UINT32 >> n) << n  # Top n bits only
    high_bits_mask = MAX_UINT32 ^ ((1 << (32 - n)) - 1)  # Clearer version
    
    bits_that_will_be_lost = (x >> (32 - n)) != 0
    
    result = (x << n) & MAX_UINT32
    
    return (result, bits_that_will_be_lost)
 
 
def safe_multiply_by_power_of_2(x: int, power: int, bit_width: int = 32) -> Tuple[int, bool]:
    """
    Multiply x by 2^power with overflow detection.
    
    This is mathematically equivalent to left shift but makes
    the semantic intent clearer.
    """
    return safe_left_shift_32(x, power) if bit_width == 32 else None
 
 
def count_leading_zeros_32(x: int) -> int:
    """Count leading zeros in a 32-bit value."""
    if x == 0:
        return 32
    
    count = 0
    mask = 0x80000000  # Start from MSB
    
    while (x & mask) == 0 and count < 32:
        count += 1
        mask >>= 1
    
    return count
 
 
def max_left_shift_without_overflow_32(x: int) -> int:
    """
    Determine the maximum left shift that won't lose bits.
    
    Returns the largest n such that (x << n) fits in 32 bits.
    """
    if x == 0:
        return 32  # Can shift by any amount; result is always 0
    
    x = x & 0xFFFFFFFF  # Ensure 32-bit
    
    # Maximum shift is determined by leading zeros
    return count_leading_zeros_32(x)
 
 
if __name__ == "__main__":
    print("Safe left shift with overflow detection:")
    print("-" * 50)
    
    test_cases = [
        (1, 31),          # 1 << 31 = 2^31, fits
        (1, 32),          # 1 << 32 = 2^32, overflow
        (0x80000000, 1),  # Already has MSB set, any shift loses it
        (0x7FFFFFFF, 1),  # Max positive, shifting doubles it
        (255, 24),        # 255 << 24 = 255 * 2^24, fits
        (255, 25),        # 255 << 25 = 255 * 2^25, overflow
    ]
    
    for x, n in test_cases:
        result, overflow = safe_left_shift_32(x, n)
        status = "OVERFLOW" if overflow else "OK"
        print(f"  0x{x:08X} << {n:2d} = 0x{result:08X} [{status}]")
    
    print("\nMaximum safe shift amounts:")
    for x in [1, 0x100, 0x10000, 0x80000000]:
        max_shift = max_left_shift_without_overflow_32(x)
        print(f"  0x{x:08X} can be shifted by at most {max_shift}")

Leading Zeros Tell the Story

The count of leading zeros in a value directly tells you the maximum left shift that won't overflow. If a 32-bit value has 5 leading zeros, you can safely left-shift it by up to 5 positions. This is why CLZ (Count Leading Zeros) is a common CPU instruction—it's fundamental to safe shift operations.

Arbitrary Precision: Python's Different World

Python's integers have arbitrary precision—they can grow to any size, limited only by available memory. This fundamentally changes how shift overflow works compared to fixed-width languages.

In Python:

Left shift never overflows; it just produces a larger number
Right shift eventually reaches 0 (or -1 for negative numbers)
There's no wraparound behavior to worry about
No undefined behavior exists for shift amounts (except negative amounts raise an error)

python_arbitrary_precision.py
Python
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def demonstrate_arbitrary_precision():
    """Show Python's arbitrary precision prevents overflow."""
    
    print("Python arbitrary precision left shift:")
    
    x = 1
    for shift in [31, 32, 63, 64, 100, 1000]:
        result = x << shift
        # No overflow! Just a very large number
        print(f"  1 << {shift:<4d} = {result}")
        if shift <= 64:
            print(f"         (binary length: {result.bit_length()} bits)")
    
    # Python can handle truly massive shifts
    huge = 1 << 10000  # A number with over 3000 decimal digits
    print(f"\n  1 << 10000 has {huge.bit_length()} bits and {len(str(huge))} decimal digits")
 
def simulate_fixed_width_overflow(x: int, n: int, width: int) -> int:
    """
    Simulate fixed-width integer overflow in Python.
    
    This is useful when porting algorithms from C or when
    you need to match fixed-width behavior exactly.
    """
    mask = (1 << width) - 1  # e.g., 0xFFFFFFFF for 32-bit
    
    # Clamp input to width
    x = x & mask
    
    # Perform shift with modular wrapping
    if n >= width:
        return 0
    
    return (x << n) & mask
 
def simulate_signed_fixed_width(x: int, n: int, width: int) -> int:
    """
    Simulate signed fixed-width arithmetic right shift in Python.
    """
    # First, interpret x as a signed value of given width
    sign_bit = 1 << (width - 1)
    max_unsigned = (1 << width) - 1
    
    x = x & max_unsigned  # Clamp to width
    
    if x & sign_bit:  # Negative in two's complement
        # x is negative; need to sign-extend
        # Python's >> on negative is arithmetic, but we have unsigned repr
        x = x - (1 << width)  # Convert to negative Python int
    
    # Now shift
    if n >= width:
        return -1 if x < 0 else 0
    
    result = x >> n
    
    # Convert back to unsigned representation
    return result & max_unsigned
 
if __name__ == "__main__":
    demonstrate_arbitrary_precision()
    
    print("\nSimulating 8-bit unsigned wraparound:")
    for x in [128, 200, 255]:
        for n in [1, 2, 4]:
            result = simulate_fixed_width_overflow(x, n, 8)
            print(f"  {x} << {n} (8-bit) = {result}")
    
    print("\nSimulating 8-bit signed arithmetic right shift:")
    for x in [0xFF, 0x80, 0x7F]:  # -1, -128, 127 in 8-bit signed
        for n in [1, 2, 4, 7]:
            result = simulate_signed_fixed_width(x, n, 8)
            # Interpret result as signed for display
            signed = result if result < 128 else result - 256
            print(f"  0x{x:02X} ({x if x < 128 else x-256:4d}) >> {n} = 0x{result:02X} ({signed:4d})")

When to simulate fixed-width behavior in Python:

Porting algorithms — When translating C code that relies on wraparound
Cryptographic code — Many crypto algorithms require exact 32-bit or 64-bit wrapping
Hardware simulation — Modeling fixed-width registers and buses
Interoperability — When Python code must produce identical results to C/Java code
Testing — Generating test vectors for fixed-width implementations

Best of Both Worlds

Python's arbitrary precision is great for prototyping and algorithms where numeric range shouldn't be a concern. For production code that must match fixed-width behavior, use the simulated functions with explicit masking. For maximum performance, consider NumPy arrays with explicit dtypes (uint32, int64, etc.).

Real-World Bugs and War Stories

Shift overflow bugs have caused real problems in production systems. Understanding these cases helps build intuition for where to look for similar issues in your own code.

Notable Shift-Related Bugs

•Java's Binary Search Bug (2006) — The implementation (low + high) / 2 overflows for large arrays. Fixed with (low + high) >>> 1. This bug existed in Java for 9 years before discovery.
•OpenSSL Heart Bleed (2014) — While not purely a shift bug, incorrect bounds checking involving size calculations led to one of the most famous security vulnerabilities.
•Video game physics glitches — Games using fixed-point arithmetic with shifts have exhibited overflow bugs causing characters to teleport or objects to disappear at extreme coordinates.
•Image processing artifacts — Incorrect handling of pixel value shifts causing color channel overflow, leading to unexpected bright spots or color inversions.
•Audio sample clipping — Audio mixing code that shifts samples without clamping, causing digital distortion when combined values exceed range.

binary_search_bug.java
Java
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public class BinarySearchBug {
    
    // THE BUG: This can overflow for large arrays
    public static int binarySearchBuggy(int[] arr, int target) {
        int low = 0;
        int high = arr.length - 1;
        
        while (low <= high) {
            // BUG: low + high can overflow if both are > Integer.MAX_VALUE / 2
            int mid = (low + high) / 2;  // Overflow here!
            
            if (arr[mid] == target) return mid;
            else if (arr[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        return -1;
    }
    
    // THE FIX: Use unsigned right shift
    public static int binarySearchFixed(int[] arr, int target) {
        int low = 0;
        int high = arr.length - 1;
        
        while (low <= high) {
            // FIX: >>> treats the (possibly negative) sum as unsigned
            int mid = (low + high) >>> 1;  // No overflow!
            
            if (arr[mid] == target) return mid;
            else if (arr[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        return -1;
    }
    
    // ALTERNATIVE FIX: Avoid overflow in the first place
    public static int binarySearchAlternative(int[] arr, int target) {
        int low = 0;
        int high = arr.length - 1;
        
        while (low <= high) {
            // ALTERNATIVE: Compute delta, then add to low
            // This never overflows because high - low <= high <= MAX_VALUE
            int mid = low + (high - low) / 2;  // Safe!
            
            if (arr[mid] == target) return mid;
            else if (arr[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        return -1;
    }
    
    public static void main(String[] args) {
        // Demonstrate the overflow
        int low = Integer.MAX_VALUE - 10;  // 2147483637
        int high = Integer.MAX_VALUE;      // 2147483647
        
        int badMid = (low + high) / 2;     // Overflow! Negative!
        int goodMid = (low + high) >>> 1;  // Correct!
        int altMid = low + (high - low) / 2;  // Also correct!
        
        System.out.println("low = " + low);
        System.out.println("high = " + high);
        System.out.println("(low + high) = " + (low + high) + " (OVERFLOW!)");
        System.out.println();
        System.out.println("Buggy mid:       " + badMid + " (WRONG!)");
        System.out.println("Fixed mid (>>>): " + goodMid);
        System.out.println("Alt mid (delta): " + altMid);
    }
}

The Lesson

Even expert programmers at major companies miss these bugs. The binary search overflow existed in Java's standard library for nearly a decade. Code reviews, testing with small values, and even formal verification didn't catch it. Only testing with arrays larger than 2^30 elements would trigger the bug—a scale rarely encountered in practice. Always consider: 'What happens at the edge of representable values?'

Summary: Surviving Shift Overflow

We've explored the dangerous territory of shift overflow and bit loss. Here are the essential survival strategies:

Key Takeaways

•Left shift loses high-order bits silently. When bits are pushed beyond the container's width, they vanish without warning.
•Right shift loses low-order bits. These represent the remainder of division and cannot be recovered after the shift.
•Shift by >= bit_width is undefined in C/C++. It wraps in Java/JavaScript but never assume specific behavior.
•Signed left shift overflow is undefined in pre-C++20. Use unsigned types for bit manipulation.
•Detect overflow before it happens. Check leading zeros, compare with maximum shiftable values, or use wider intermediate types.
•Python has no overflow. But when matching fixed-width behavior, explicit masking is needed.
•Test at boundaries. Edge cases with maximum values, negative numbers, and extreme shift amounts reveal bugs that normal testing misses.

Putting it all together:

You now have a complete understanding of bit shifting operations—left shift for multiplication, right shift for division, the logical vs. arithmetic distinction, and the pitfalls of overflow and bit loss. This knowledge forms the foundation for all bit manipulation techniques.

With these four pages complete, you're ready to apply bit shifting in real algorithms: creating bitmasks, packing/unpacking data, fast arithmetic, and all the clever tricks that make low-level code efficient. The next module will build on this foundation with more advanced bit manipulation patterns.

Module Complete

Congratulations! You've mastered bit shifting—the fundamental operation underlying countless optimizations in systems programming, embedded development, graphics, cryptography, and algorithm design. You understand not just how shifts work, but why they work, when they fail, and how to use them safely. This knowledge separates engineers who merely use bit operations from those who truly understand them.