printf statement inside the if condition

Question

Can anyone explain how does this work the output is A3 but how come it print 3

#include <stdio.h>

int main() {
    int i;
    if(printf("A"))
        i=3;
    else
        i=5;
    printf("%d",i);
}

Answer 1

printf() returns the number of characters upon success and negative values on failure.

Therefore, if printf("A") succeeds, it will return 1 .

In C, values other than 0 is treated as true, so i=3; is executed.

Answer 2

Let's check the flow:

    int i;            --> i has indeterminate value
    if(printf("A"))   --> prints A and returns 1, so the condition is TRUE (see note)
        i=3;          --> This statement is executed
    else              --> this condition is skipped
        i=5;          --> so this does not execute
    printf("%d",i);   --> prints the value of i which is 3.

Final print is A3 .

That said, if no conversion specification is needed, instead of using printf() , one should use puts() or fputs() .

Note:

From man printf()

Return value Upon successful return, these functions return the number of characters printed (excluding the null byte used to end output to strings).

Answer 3

the trick to understand this behav is here and the return val od printf

returns the number of characters upon success and negative values on failure. this here:

if(printf("A"))

can be read as

int r = printf("A");// at this point r ==1

if(1) //this here is true so i is assigned to 3