Why would you return(0*ap++)?

Question

It's K&R-C and here is the whole code: http://v6shell.org/history/if.c

Look at this function:

char *nxtarg() {
    if (ap>ac) return(0*ap++);
    return(av[ap++]);
}

1.Question: Why return (0*ap++)? Okay you want to return 0 and increase ap by 1. But why like this? Is it faster than if (ap>ac) {ap++; return 0;}?

2.Question: The return type of nxtarg has to be char*, why can you return 0, an integer?

Answer 1

This is a little trick to squeeze the increment into a statement that returns zero. It is logically equivalent to a conditional

if (ap > ac) {
    ap++;
    return 0;
}

or even better with a comma operator:

return (ap++, (char *)0); // Thanks, Jonathan Leffler

Note that since zero in your example is not a compile-time constant, the expression needs a cast to be compliant with the standard:

if (ap>ac) return (char*)(0*ap++);

As far as returning an integer zero goes, it is considered to be equal to NULL pointer when used in pointer context.

Answer 2

The ++ operator will increment the value, and return the value before it was incremented. If the function was supposed to return a pointer, you can return zero or NULL, to indicate a NULL pointer.