[FIXED] How does the following code works printf("%c")?

Issue

I wanted to know how the following program is working?

#include <stdio.h>
int main(void) {
    while(1){
        if(printf("%d",printf("%c")))
        break;
        else
        continue;
    }
    return 0;
}

I did not know how the part printf("%c") is working and therefore the whole program.I am aware of writing something like printf("%c", 'a'); like that but how is it working without providing the character to be printed? My question is what does the following program prints and how does it prints so?

I have tried to run the program, sometimes it prints nothing, but sometimes it is printing some random character followed by 1. I am not able to get how it is working, can someone please explain what is going behind the code and how it is printing that random characters, and why there is one at the end?

Here are some output I am getting

Solution

A cool wrong program you have.

printf("%c") attempts to print a single character that is supposed to be the second parameter. However, since you have never passed the second parameter, the function prints whatever is in the register that was supposed to have the second parameter. In other words, some random character. However, it prints one character and returns 1: the number of characters printed.

That 1 is in turn printed by printf("%d",printf("%c")). Now you have a random character followed by 1, and since the outer printf also prints one character, it returns 1.

Finally, if(printf("%d",printf("%c"))) interprets that later 1 as true and breaks the loop.

Answered By - DYZ

Answer Checked By - Pedro (PHPFixing Volunteer)