Why Does (1 >> 32) Return 1 in C : A Deep Dive into Right-Shift Operator Behavior and Undefined Behavior?

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Curiosity surrounding the Right-Shift Operator (1 >> 32)

When utilizing the right-shift operator (>>) in C code, it's often assumed that shifting by a value greater than or equal to the width of the operand will result in zero. However, as demonstrated in the provided code snippet, this assumption can lead to unexpected behavior.

The function foo(a, b) shifts integer a by b bits. When called with arguments (1, 32), it surprisingly returns 1 instead of the expected 0. This behavior stems from the compiler treating the constant expression 1 >> 32 at compile-time, where it evaluates to 0 due to the undefined behavior described in the C standard.

In contrast, the bar(a, b) function operates on a 64-bit unsigned integer and correctly returns 0 when shifting by 32 bits. This is because 64 is greater than 32, guaranteeing that the shift will produce 0. However, even for bar, shifting by 64 bits may still return 1.

Elucidating this behavior further, the hardware implementation of the shift operation masks the shift count to 5 or 6 bits (depending on the architecture), effectively truncating any shift count greater than or equal to 32 or 63. Therefore, the logical right shift (LSR) on certain architectures ensures that shifts of ≥32 will always produce zero.

This highlights the non-portable nature of shifting a 32-bit integer by ≥32, as the result can vary depending on the underlying hardware implementation.

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