how %c prints value in a C program?

Question

The following program yields 12480 as the output.

#include<stdio.h>

int main()
{
    char c=48;
    int i, mask=01;
    for(i=1; i<=5; i++)
    {
        printf("%c", c|mask);
        mask = mask<<1;
    }
    return 0;
}

Now, my question is, how "%c" prints the integer value 1, 2, 4, 8, 0 after every loop. It should print a character as a value. If i simply use the following program,

#include<stdio.h>

int main()
{
    char c=48;
    int i, mask=01;
    printf("%c",c); 
    return 0;
}

it prints 0 but when i change the identifier %c to %d it prints 48 . Can anyone please tell me how is this going!?

Answer 1

If you use %c, c prints the corresponding ASCII key for the integer value.

Binary of 48 is 110000. Binary of 1 is 000001.

You or them, 110000 | 000001 gives 110001 which is equivalent to 49 in decimal base 10.

According to the ASCII table, corresponding ascii values for 49, 50, 51, etc are '1', '2', '3', etc.