Weird behavior of right shift operator (1 >> 32)

Question

I recently faced a strange behavior using the right-shift operator.

The following program:

#include <cstdio> #include <cstdlib> #include <iostream> #include <stdint.h>  int foo(int a, int b) {    return a >> b; }  int bar(uint64_t a, int b) {    return a >> b; }  int main(int argc, char** argv) {     std::cout << "foo(1, 32): " << foo(1, 32) << std::endl;     std::cout << "bar(1, 32): " << bar(1, 32) << std::endl;     std::cout << "1 >> 32: " << (1 >> 32) << std::endl; //warning here     std::cout << "(int)1 >> (int)32: " << ((int)1 >> (int)32) << std::endl; //warning here      return EXIT_SUCCESS; } 

Outputs:

foo(1, 32): 1 // Should be 0 (but I guess I'm missing something) bar(1, 32): 0 1 >> 32: 0 (int)1 >> (int)32: 0 

What happens with the foo() function ? I understand that the only difference between what it does and the last 2 lines, is that the last two lines are evaluated at compile time. And why does it "work" if I use a 64 bits integer ?

Any lights regarding this will be greatly appreciated !

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Surely related, here is what g++ gives:

> g++ -o test test.cpp test.cpp: In function 'int main(int, char**)': test.cpp:20:36: warning: right shift count >= width of type test.cpp:21:56: warning: right shift count >= width of type 

Answer 1

It's likely the CPU is actually computing

a >> (b % 32) 

in foo; meanwhile, the 1 >> 32 is a constant expression, so the compiler will fold the constant at compile-time, which somehow gives 0.

Since the standard (C++98 §5.8/1) states that

The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand.

there is no contradiction having foo(1,32) and 1>>32 giving different results.

 

On the other hand, in bar you provided a 64-bit unsigned value, as 64 > 32 it is guaranteed the result must be 1 / 2³² = 0. Nevertheless, if you write

bar(1, 64); 

you may still get 1.

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Edit: The logical right shift (SHR) behaves like a >> (b % 32/64) on x86/x86-64 (Intel #253667, Page 4-404):

The destination operand can be a register or a memory location. The count operand can be an immediate value or the CL register. The count is masked to 5 bits (or 6 bits if in 64-bit mode and REX.W is used). The count range is limited to 0 to 31 (or 63 if 64-bit mode and REX.W is used). A special opcode encoding is provided for a count of 1.

However, on ARM (armv6&7, at least), the logical right-shift (LSR) is implemented as (ARMISA Page A2-6)

(bits(N), bit) LSR_C(bits(N) x, integer shift)     assert shift > 0;     extended_x = ZeroExtend(x, shift+N);     result = extended_x<shift+N-1:shift>;     carry_out = extended_x<shift-1>;     return (result, carry_out); 

where (ARMISA Page AppxB-13)

ZeroExtend(x,i) = Replicate('0', i-Len(x)) : x 

This guarantees a right shift of ≥32 will produce zero. For example, when this code is run on the iPhone, foo(1,32) will give 0.

These shows shifting a 32-bit integer by ≥32 is non-portable.