Bit toggling confusion in C

Question

Hi I have written a program to toggle bits in char array. I found that when I am toggling the 7th bit I am getting wrong answer. Please help me.

int main() {
    int n,c;
    char dummy;
    scanf("%i", &n);
    char a[13];
    memset(a,0x00,13);
    for(int a_i = 0; a_i < n; a_i++){
       scanf("%d",&c);
        dummy = c%8;
        a[c/8] ^= (1<<dummy);
    }
    printf("\n");
    for (int _i=0;_i<13;_i++)printf("%x ",a[_i]);
//    int result = lonelyinteger(n, a);
//    printf("%d\n", result);
    return 0;
}

Input:
9
4 9 95 93 57 4 57 93 9

Output:
0 0 0 0 0 0 0 0 0 0 0 **ffffff80** 0

I need the output as 80 only not ffffff80. Please help me with this.

`printf("%x ", a[_i]);` --> `printf("%x ", (unsigned char)a[_i]);` — BLUEPIXY, Oct 11 '17 at 03:46
it is sign extending the byte to a 32 bit number so (unsigned int)(a[-i]) may/should fix that. Or dummy=a[_i]; then printf the dummy variable — old_timer, Oct 11 '17 at 03:46
@old_timer Casting to `unsigned int` does not work as expected. — BLUEPIXY, Oct 11 '17 at 03:48
@BLUEPIXY Thanks for the help. unsigend char is working. I have declared the array as unsigned char a[13]. — Damodharan_C, Oct 11 '17 at 03:49

score -1 · Answer 1 · answered Oct 11 '17 at 04:59

-1

Remember that it is not only the signed-ness of the array type, but the signed-ness of the integer literal that you are shifting that matters. Make sure both are unsigned.

x ^= 1U << n;

answered Oct 11 '17 at 04:59

Dúthomhas

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This is good general advice; however `dummy` never exceeds `7` in OP's code, and there is no difference between `x ^= (1 << 7)` and `x ^= (1u << 7)`. The problem is more to do with the declared type of `x` – M.M Oct 11 '17 at 05:16

Bit toggling confusion in C

1 Answers1