For loop with printf as arguments

Question

I can't understand why the following code outputs 10 . What I understand is that !printf("0") means !0 , which is TRUE . So why doesn't the code print "Sachin"

#include <stdio.h>

int main() {
    for (printf("1"); !printf("0"); printf("2"))
        printf("Sachin");
    return 0;
}

Output

10

Answer 1

let's analyze this side-effect loop statement:

for(printf("1"); !printf("0"); printf("2"))

• The first statement is executed, always (init condition), yieiding 1
• Then the condition is tested: !printf("0") prints 0 , then since printf returns 1 because it just prints 1 character, the negation returns 0 and the loop is never entered because the condition is false right from the start. So neither 2 or Sachin are printed.

Of course, this code isn't practical, almost unreadable. So don't ever do things like this ( puts("10"); is a good alternative for instance).

more on the return value of printf (that is often ignored):

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

(from https://linux.die.net/man/3/printf )

Answer 2

If you look at the man printf reference on google, you'll see that this function returns the number of written bytes.

Here your condition is !printf("0") , in other words : "as long as the return of printf is not existing (or equal 0), do something. But you print the character '0' so printf actually return 1 so your condition is false.

Now WHY it prints 10 :

• The first printf("1") prints 1.
• Your condition is tested at least once, so the second printf("0") occurs one time (it prints 0)

Answer 3

printf("1")

prints 1 and it return number of characters which is 1

printf("0")

prints 0 and it return number of characters which is 1

!1 means !(true) = false so execution will stop and you will see 10 as output.