问题描述

我一定是疯了，但是我机器上的 gcc 4.7.3 给出了最荒谬的结果.这是我正在测试的确切代码:

I must be absolutely crazy here, but gcc 4.7.3 on my machine is giving the most absurd result. Here is the exact code that I'm testing:

#include <iostream>

using namespace std;

int main(){
  unsigned int b = 100000;
  cout << (b>>b) << endl;
  b = b >> b;
  cout << b << endl;
  b >>= b;
  cout << b << endl;
  return 0;
}

现在，任何自行右移的数字都应该导致 0 (n/(2^n) == 0 与 整数除法、n>1 和 positive/unsigned)，但不知何故这是我的输出:

Now, any number that's right shifted by itself should result in 0 (n/(2^n) == 0 with integer divide, n>1, and positive/unsigned), but somehow here is my output:

100000
100000
100000

我疯了吗?可能会发生什么?

Am I crazy? What could possibly be going on?

推荐答案

在 C++ 中与在 C 中一样，移位仅限于移位值的大小(以位为单位).例如，如果 unsigned int 是 32 位，那么超过 31 位的移位是未定义的.

In C++ as in C, shifts are limited to the size (in bits) of the value shifted. For instance, if unsigned int is 32 bits, then a shift of more than 31 is undefined.

在实践中，一个常见的结果是使用移位量的最低5位，忽略高位；这是因为编译器生成了一条完全符合此要求的机器指令(例如 x86 上的 SHR).

In practice, a common result is that the 5 least-significant bits of the shift amount are used and the higher order bits are ignored; this is due to the compiler producing a machine instruction which does exactly that (eg SHR on x86).

在这种情况下，移位值是 100000(十进制)，恰好是二进制的 11000011010100000 - 低 5 位为零.所以，你实际上得到了一个 0 的转变.但是你不应该依赖这个；从技术上讲，您看到的是未定义行为.

In this case, the shift value is 100000 (decimal) which happens to be 11000011010100000 in binary - the lower 5 bits are zero. So, you're effectively getting a shift by 0. You shouldn't rely on this however; technically, what you're seeing is undefined behaviour.

参考文献:

对于 C，N1570 6.5 节.7:

For C, N1570 section 6.5.7:

如果右操作数的值为负或大于或等于提升的左操作数的宽度，行为是未定义.

If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.

对于 C++，N3690第 5.8 节[expr.shift]":

For C++, N3690 section 5.8 "[expr.shift]":

如果右操作数为负数或更大，则行为未定义大于或等于提升的左操作数的位长度.

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.

N1570 是一个草案，与已发布的 ISO C11 标准几乎相同；该条款自 1989 年 ANSI C 标准以来几乎相同.

N1570 is a draft, nearly identical to the released ISO C11 standard; this clause has been pretty much the same since the 1989 ANSI C standard.

N3690 是 C++ 标准的最新草案；我不确定它是否是最好用的，但同样，这个条款没有改变.

N3690 is a recent draft of the C++ standard; I'm not sure whether it's the best one to use, but again, this clause hasn't changed.

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