CODEWITHSUNDEEP
c
Rebooting
Rebooting

Reputation: 2938

how %c prints value in a C program?

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!?

Upvotes: 7

Views: 33622

Answers (2)

shadyabhi
shadyabhi

Reputation: 17244

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.

Upvotes: 8

NPE
NPE

Reputation: 500933

It actually prints out the characters '1', '2', '4' etc.

The numeric value of c|mask gets interpreted as an ASCII code. The ASCII code of '0' is 48.

To make the code a little clearer, you could change

char c=48;

to

char c='0';

The two forms are equivalent.

Upvotes: 3

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