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?

Answer 1

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.

Answer 2

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) { ... }