C++ uint64_t 按位与检查偶数

Question

我有以下代码可以简单地检查 uint64_t 是否为偶数，我打算使用按位 AND 运算来检查但它似乎不起作用。

这是我认为首先可用的代码：

int n;
scanf("%d",&n);
for(int i = 0; i < n; i++){
    uint64_t s,d;
    scanf("%llu %llu",&s,&d);
    //try for x
    uint64_t x;
    bool stop = false;
    x = s + d;
    printf("%llu",x&1ULL); \ This prints 0 when the number is even but
    if(x&1ULL==0ULL){ \ This check always returns false
        printf("%llu",x);
        x/= 2;

如果数字是奇数或偶数，此代码总是打印出 0 或 1，但 if 语句总是 returns false。我究竟做错了什么？谢谢

Answer 1

x&1ULL==0ULL 等价于 x&(1ULL==0ULL)。你需要 (x&1ULL)==0ULL.

Answer 2

#include "stdafx.h"
#include <stdio.h>
#include <windows.h>
#include <stdint.h>



int main()
{
    int n;
    scanf_s("%d", &n);
    for (int i = 0; i < n; i++) {
        uint16_t  s, d;
        scanf_s("%llu %llu", &s, &d);
        //try for x
        uint16_t x;
        bool stop = false;
        x = s + d;
        printf("%llu", x & 1ULL);
        if ((x & 1ULL) == 0ULL) {

            printf("%llu", x);

            x /= 2;
        }
    }
    return 0;
}

我认为这对我来说应该有用它有用

C++ uint64_t 按位与检查偶数

C++ uint64_t bitwise AND check for even number

c

bit-manipulation

bitwise-operators

uint64

bitwise-and