Question

I have following c code:

#define ASSERT_ACTIVE(active) do { if (!active) return errno = 6, -1; } while (0);
#define errno (*_errno())

int someCrazyFunc (bool active) {
    ASSERT_INACTIVE (active);
    ...
}

As far as I know a #define will simply place replacement text in place of the specified identifier.

I like to know:

• What does return errno = 6, -1; means? is that returns two values in one return statement?
• What is the meaning of replacement code (*_errno()) = 6

Answer 1

There isn’t a second value – a return statement returns exactly one value. In the statement:

return errno = 6, -1;

The return value is the result of the expression errno = 6, -1. This is an expression using the comma operator – it is parsed as (errno = 6), -1, which evaluates to -1 and assigns 6 to errno as a side-effect. So this means that it’s equivalent to the two statements:

errno = 6;
return -1;

Assuming that _errno() is a function returning a pointer – for example it has a return type of int * – then the expression (*_errno()) = 6 assigns the value 6 to the object pointed to by the return value of the function. It would be equivalent to code similar to:

int *p = _errno();
*p = 6;

errno is often defined like this in order to give each thread in a multi-threaded implementation its own errno. The function _errno() in this case would return a pointer to the current thread’s errno variable.