What does `return x ? : 1` mean in C language?

Question

I have come across the following code in the Linux Kernel source (2.6.32).

do_wait_for_common(struct completion *x, long timeout, int state)
{
        if (!x->done) {
        /* some code here */
        }
        x->done--;
        return timeout ?: 1; <--- What it returns?
}

To understand the behavior, I have manually tried the following code

#include <stdio.h>
int f(int x)
{
        return x?:1;
}
int main()
{
        printf("f %d\n", f(0));
        printf("f %d\n", f(1));
        return 0;
}

And got the following output

f 1
f 1

And when I change it to

int f(int x)
{
        return x?:2;
}

I am getting

f 2
f 1

I just want to know whether this behavior (return 1 if nothing mentioned) is mentioned in the standard.

Answer 1

This behavior is not mentioned in C standard. Its a GCC extension.

The middle operand in a conditional expression may be omitted. Then if the first operand is nonzero, its value is the value of the conditional expression.

Therefore, the expression

 x ? : y

has the value of x if that is nonzero; otherwise, the value of y.

This example is perfectly equivalent to

 x ? x : y  

In this simple case, the ability to omit the middle operand is not especially useful. When it becomes useful is when the first operand does, or may (if it is a macro argument), contain a side effect. Then repeating the operand in the middle would perform the side effect twice. Omitting the middle operand uses the value already computed without the undesirable effects of recomputing it.

Answer 2

It is a GCC extension. x?:2 is the same as x?x:2 (with the then part evaluated once)

Answer 3

This is a GNU C extension of the normal ternary operator. X ?: Y is the same as X ? X : Y except that X is only evaluated once.

Answer 4

As per GCC extension

The middle operand in a conditional expression may be omitted

return x?:1; is a short-hand method to write return x?x :1;