Reputation: 1965

Unusual behavior of ternary operator?

i was just playing around with the ternary operator in my c class today. And found this odd behavior.

#include <stdio.h>
#include <stdbool.h>
main()
{
        int x='0' ? 1 : 2;
        printf("%i",x);
}

returns 1 as expected.But

#include <stdio.h>
#include <stdbool.h>
main()
{
        int x='0'==true ? 1 : 2;
        printf("%i",x);
}

returns 2 while i expect it to return 1.

Upvotes: 1

Answers (6)

ruakh

Reputation: 183251

That's because (assuming ASCII) '0' represents the integer 0x30, i.e. 48, and true represents the integer 1. So they're not equal.

In C, any nonzero value is considered true, but true itself is 1, and 1 is what you get from any built-in Boolean test (for example, 0 == 0 is 1).

Upvotes: 1

ouah

Reputation: 145829

Maybe you are confusing '\0' and '0'

The value of the character constant

'\0'

is always 0 in C.

The value of the character constant

'0'

is implementation defined and depends on the character set. For ASCII, it is 0x30.

Also note that the macro true is defined to the value 1 and not 0.

Upvotes: 1

moonshadow

Reputation: 89055

'0' does not equal true, so the result of '0'==true is 0 (false). This is assigned to x and passed to the ternary operator, giving the result you see.

If you intended something else, you should use brackets to clarify the order of precedence you want, or break your code up into multiple statements.

Upvotes: 0

unwind

Reputation: 399763

The value of '0' is not zero, it is whatever integer value encodes the digit '0' on your system. Typically 48 (in encodings borring from ASCII), which is then not equal to true when interpreted as an integer, which is 1.

So the first of your code lines is equivalent to