为什么"int>> 32`并不总是为零吗? [英] Why is `int >> 32` not always zero?

问题描述

有人可以解释一下为什么在C/C ++中对某个4字节整数进行右32位按位移位可能不会返回零吗?为什么它取决于编译器的-O选项?

Can someone explain me why right 32 bitwise shift of some 4 byte integer number may return not zero in C/C++ ? Why does it depend on -O option of the compiler?

例如，这段代码在gcc 4.8.3中为-O0提供了45个，为-O3提供了0:

For example this code gives 45 with -O0 and 0 with -O3 options in gcc 4.8.3:

unsigned int x = 45; // 4 bytes
x = x >> 32;
printf("%u\n", x);

为什么会这样?

推荐答案

因为它是未定义的行为，所以:[expr.shift]说

Because it is undefined behavior: [expr.shift] says

如果右操作数为负或大于或等于提升后的左操作数的位长度，则该行为是不确定的.

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.

对于特定的未定义行为，我想它是这样的:

As for the specific undefined behavior, I imagine it is as follows:

• 使用-O0，它可以编译为实际执行机器代码的右移，并且在某些机器上(例如，我相信x86就是这样)，当将32移位32时，移位函数仅查看移位量的低5位.位字；移32等于移0.
• 使用-O3，编译器自己计算常量，然后将0放入程序中，而不是让它进行计算.

• With -O0, it compiled to actually perform a right shift in machine code, and on some machines (e.g. I believe x86 is such), shift functions only look at the low 5 bits of the shift amount when shifting a 32-bit word; shifting by 32 is the same as shifting by 0.
• With -O3, the compiler computed the constant itself, and just put 0 into the program rather than having it do a calculation.

您可以检查程序集的输出以查看我的预测是否正确.

You can check the assembly output to see if my prediction is right.

这篇关于为什么"int>> 32`并不总是为零吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！