C how to mask bits into char array

Question

I have a char[16] array and i'm getting input from the user: Input for example- 15, 21 ,23, -1

I need to set the bit value to '1' for the place 15,21 and 23. -1 will finish the program.

Every char[16] array represent values from 0-127, that represents bits. I'm having problem entering '1' into 15,21 and 23 cells.

Here is my program

int temp;
char A[16];
/*Sets all the cells values to o*/
memset(A, 0, 16*sizeof(char));
While (int != -1)
{
    scanf("Enter values from the user:%d", val");
    div = (temp/8);
    mod = (temp%8);
    A[div] |= (mod<<=1);
}

The problem that it's not setting cell 15,21 and 23 values to '1'.

Answer 1

Use this to set the right bit:

A[div] |= (1<<mod);

Related question: How do you set, clear, and toggle a single bit?

Full code example:

#include <iostream>

int main() {
    unsigned char A[16];
    memset(A, 0, sizeof(A));
    int t;
    std::cin >> t;
    while (t != -1)
    {
        int div = (t/8);
        int mod = (t%8);
        A[div] |= (1<<mod);
        std::cin >> t;
    }
    for(int i = 0; i < 16; ++i) {
        std::cout << (int)A[i] << " ";
    }
    std::cout << std::endl;
    return 0;
}

Answer 2

bit-fields are not defined for char(if u use char use unsigned.. ) , use unsigned int. Or the C99 boolean type. https://stackoverflow.com/a/3971334/1419494