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

Perhaps you need [a `printf` reference](https://en.cppreference.com/w/c/io/fprintf) to see what it *returns*? — Some programmer dude, Sep 11 '18 at 07:53
`printf` returns the number of characters printed. It can print `1`, `10` or nothing at all, depending on after which character it does fail. — Antti Haapala -- Слава Україні, Sep 11 '18 at 07:54
!printf("0") returns false. printf itself returns the number of chars printed if successful. check any c reference for details. — Zzyzx, Sep 11 '18 at 07:58
`!printf("0")` means `!0`... No. Given the result of `printf` it means `!1` which is `FALSE` — Gerhardh, Sep 11 '18 at 08:37

Jean-François Fabre · Answer 1 · 2018-09-11T11:56:38.203

8

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)

edited Sep 11 '18 at 11:56

answered Sep 11 '18 at 07:58

Jean-François Fabre

137,073
23
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Meaning no offence sir, this code may not be practical but to me it's experimental which is important to understand the things better. However, I thank you for your assistance. – Debmalya Panday Sep 11 '18 at 10:41
no offence taken. I added that bit in my answer so everyone understands that if you want to print `10` you just do `printf("10");` – Jean-François Fabre Sep 11 '18 at 11:55
1

And only write `puts("10");` if you also require a newline. ;-) – Toby Speight Sep 11 '18 at 14:58
aaah right I wanted to avoid `printf` when no formatting was needed :) – Jean-François Fabre Sep 11 '18 at 14:59

Malorke · Answer 2 · 2018-09-11T09:01:01.247

3

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)

edited Sep 11 '18 at 09:01

answered Sep 11 '18 at 08:01

Malorke

53
1
6

2

good first answer. Small piece of advice: don't tell people to "look on google". Google for them and include the link. It's better. – Jean-François Fabre Sep 11 '18 at 08:51
Thanks for the assistance. – Debmalya Panday Sep 11 '18 at 10:43

Milan Shah · Answer 3 · 2018-09-12T06:25:11.530

0

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.

edited Sep 12 '18 at 06:25

answered Sep 11 '18 at 08:51

Milan Shah

111
1
10

Understood. Thanks for the assistance. – Debmalya Panday Sep 11 '18 at 10:43

For loop with printf as arguments

3 Answers3