Output behaviour of relational operators when used in printf - C Language

Question

This is the code that i was trying :

#include <stdio.h>
main(){
    int k = 35;
    int x = (k==35);
    printf("%d %d %d %d", (k==35), k=50, (k>40), x); // gives output => 0 50 0 1
    printf("\n%s",k==35); // gives output => (null)
    return 0;
}

I was wondering that a relational check returns Boolean true or false. Even when converted to integer type it should become 0 for false and 1 for true. What am i missing?

You're not missing anything. Why do you think you are? – Barmar Jun 19 '14 at 10:31 — Barmar, Jun 19 '14 at 10:31

Basile Starynkevitch · Accepted Answer · 2014-06-19T10:37:29.680

Actually your k=50 with a single = is an assignment operator; since it appears in a list of arguments, the order of evaluation of arguments is undefined, and you get stuck by undefined behavior.

The compiler is free to pass (k>40) before or after doing k=50

BTW, with GCC version 4.9 on Debian/Sid, when compiling your code with

gcc -Wall danglingcruze.c -o danglingcruze

I am warned by the compiler:

danglingcruze.c:2:1: warning: return type defaults to ‘int’ [-Wreturn-type]
 main(){
 ^
danglingcruze.c: In function ‘main’:
danglingcruze.c:5:37: warning: operation on ‘k’ may be undefined [-Wsequence-point]
     printf("%d %d %d %d", (k==35), k=50, (k>40), x); // gives output => 0 50 0 1
                                     ^
danglingcruze.c:5:37: warning: operation on ‘k’ may be undefined [-Wsequence-point]
danglingcruze.c:6:5: warning: format ‘%s’ expects argument of type ‘char *’, 
        but argument 2 has type ‘int’ [-Wformat=]
     printf("\n%s",k==35); // gives output => (null)
     ^

Yes that outputs 50 and that is okay for me. The problem is when i compare even if it is true it returns 0. — Dangling Cruze, Jun 19 '14 at 10:31

score 2 · Answer 2 · answered Jun 19 '14 at 10:30

Order of evaluation of parameters in a function is not guaranteed. You are modifying k in one of the parameter and using its value in other parameters. That's why it invokes undefined behavior. An example of similar case

int i = 1;
printf("%d %d\n", i++, ++i); // Undefined behavior.

score 0 · Answer 3 · answered Jun 19 '14 at 10:30

0

#include <stdio.h>
main(){
    int k = 35;
    int x = (k==35); // x =1
    printf("%d %d %d %d", (k==35), k=50, (k>40), x); // gives output => 0 50 0 1
    // k=50 changed your value
    printf("\n%s",k==35); // gives output => (null)
    // => printf %s + 0 => null 
    return 0;
}

answered Jun 19 '14 at 10:30

Vincent Guerci

14,379
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50
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I don't know but for me `printf("\n%d",k==35);` it is giving 0 as output. – Dangling Cruze Jun 19 '14 at 10:32
Yes, why 0? why not 1? – Dangling Cruze Jun 19 '14 at 10:33
Isn't it an undefined behavior then? I am not able to understand what it is you're trying to convey? – Dangling Cruze Jun 19 '14 at 10:37
k==35 it will return 1 but `printf` takes address of an variable as an argument. And 1 is not a valid address. – sutirtha Jun 19 '14 at 10:38
@DanglingCruze `k == 35` returns 0 because you did `k = 50`. – Barmar Jun 19 '14 at 10:39
1

@sutirtha The %d format expects an integer as an argument, not an address. – Barmar Jun 19 '14 at 10:39
@Barmar I understand that i'm assigning the value inside function parameter which is causing the undefined behavior. Thank you :) – Dangling Cruze Jun 19 '14 at 10:42

Output behaviour of relational operators when used in printf - C Language

3 Answers3