C++ uint64_t bitwise AND check for even number

Question

I have the following code that simply checks if a uint64_t is even, I intended on using a bitwise AND operation to check but it doesn't seem to be working.

This is the code I thought would work first:

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;

This code always prints out 0 or 1 if the the number is odd or even but the if statement always returns false. What am I doing wrong? Thanks

If you're not certain of the operator precedence rules, put in the extra parentheses. Also if you are certain. — Alan Stokes, Mar 09 '16 at 23:26
If this is C++, why not use `std::cin`? This is 100% C code. — tadman, Mar 09 '16 at 23:26
@tadman Some like iostream, some hate it and some love-hate it. Nothing wrong with using the old apis. — BitTickler, Mar 09 '16 at 23:28
@tadman in this case this program is for a programming competition so I use printf,scanf and bitwise operations as they are much faster, You are correct though, in general practice I would use cin — Eoin Davey, Mar 09 '16 at 23:37
@AlanStokes If you're not certain of the operator precedence rules, look them up. — wildplasser, Mar 09 '16 at 23:45
@EoinDavey Faster in what sense? How many billions of operations of this function are you doing? — tadman, Mar 10 '16 at 02:59

score 5 · Accepted Answer · answered Mar 09 '16 at 23:29

5

x&1ULL==0ULL is equivalent to x&(1ULL==0ULL). You need (x&1ULL)==0ULL.

answered Mar 09 '16 at 23:29

Alan Stokes

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score 0 · Answer 2 · answered Mar 09 '16 at 23:45

#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;
}

I think this should work anyway for me it works

C++ uint64_t bitwise AND check for even number

2 Answers2