Operator precedence and ternary operator

Question

I have a problem in C.

#include<stdio.h>
int main()
{
    int a = 10, b = 0, c = 7;
    if (a ? b : c == 0)
        printf("1");
    else if (c = c || a && b)
        printf("2");
    return 0;
}

This code prints 2 but I think a?b:c returns b=0 and 0==0 returns 1. Can you explain the code?

Why don't you just look up the operator precedence table? Such kinds of questions are useless otherwise. — Blagovest Buyukliev, Apr 04 '12 at 08:06
If you know this is about precedence, why don't use the [precedence order](http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence) to figure out where to put parentheses? — Oliver Charlesworth, Apr 04 '12 at 08:06

score 8 · Answer 1 · answered Apr 04 '12 at 08:08

8

The conditional operator (?:) has one of the lowest precedences. In particular it is lower than ==. Your statement means this:

if(a ? b : (c == 0)) { ... }

Not this:

if((a ? b : c) == 0) { ... }

answered Apr 04 '12 at 08:08

Mark Byers

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score 4 · Accepted Answer · edited Feb 09 '14 at 04:09

Your conditions are not properly written.

In the first if-statement:

  if (a ? b : c == 0)

if you put the values, then it becomes

if(10 ? 0 : 7 == 0)

means, it will always return 0.

That's why control goes to the else part and there, it becomes

else if (7 = 7 || 10 && 0)

since you used the "=" operator here (c = c), it will be always true, therefore it prints "2".

Now you want that code should return "1", then change your if statement in this way.

 if( (a ? b:c) == 0){...}

because "==" operator has higher precedence than ternary operator.

Operator precedence and ternary operator

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