'Return' as a function in an array of function pointers

Question

So here's what I'm working with.

{
    int Test;
    void (*functions[5])() = {
        return,
        one,
        two,
        three,
        four
    };

    for (Test = takeInput(5);; Test = takeInput(5)){
        cls();
        Print(L"0)  Abort\n");
        Print(L"1)  One\n");
        Print(L"2)  Two\n");
        Print(L"3)  Three\n");
        Print(L"4)  Four\n");
        Print(L"Enter Selection: ");
        (*functions[Test])();
    }
}

That's the gist of what I'm trying to do. This is sort of a menu system for some context (and takeInput(5), will pause for input, stopping the loop until someone inputs a number).

What the problem is, is that using return in that way is apparently wrong. I figured that seeing as how (i thought) return is a function, i can't see why it wouldn't be able to be put into a function pointer array in order to break out of the menu. I suppose I could make a dummy function that basically just returns true when called, and check for that (the problem with that being that all the other functions return void). Even if that worked, I would rather have a solution that is as self contained as the above, non-working example.

Is there a nice, clean, hopefully short way of doing this that is better than anything I've thought of yet?

Answer 1

return is not a function, it has no address and therefore you can't have it in the array of pointers to functions.

Further, Test appears uninitialized when you use it the first couple of times:

for (Test;

and

(*functions[Test])();

If you want to break out of the for loop, you have to use either break or return or an appropriate condition between the two semicolons in for(;condition;) .

A function call will not break the loop. Unless, it's something exotic like abort() , exit() or longjmp() .

Answer 2

我将用NULL替换函数数组中的return并在执行函数之前测试null，当函数指针为NULL时返回。