64位移位问题 [英] 64bit shift problem

问题描述

为什么这段代码最后一个元素不是写0而是18446744073709551615?(用 g++ 编译)

Why this code does not write 0 as a last element but 18446744073709551615? (compiled with g++)

#include <iostream>

using namespace std;
int main(){
    unsigned long long x = (unsigned long long) (-1);
    for(int i=0; i <= 64; i++)
        cout << i << " " << (x >> i) << endl;
    cout << (x >> 64) << endl;
    return 0;
}

推荐答案

当你移动一个值的位数超过字长时，它通常会被mod word-size移动代码>.基本上，将其移位 64 意味着移位 0 位，这等于根本不移位.但是您不应该依赖它，因为它不是由标准定义的，并且在不同的架构上可能会有所不同.

When you shift a value by more bits than word size, it usually gets shifted by mod word-size. Basically, shifting it by 64 means shifting by 0 bits which is equal to no shifting at all. You shouldn't rely on this though as it's not defined by the standard and it can be different on different architectures.

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