Puzzle for Today

Mar 14, 2005 - c++

Can you make a function in C that returns itself?

int main() {
    assert(self == self()()()()()()()()());
    return 0;
}

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Imported Comments [?]

sufianrhazi on Mar 15, 2005

[HTML_REMOVED]no?[HTML_REMOVED]

But seriously, if you managed to pull this off do tell, do tell.

lordgalbalan on Mar 15, 2005

That seems to go against the rules of logic, in a Godelian kind of way...?

Although by use of recursion, of course. In the strictest sense. But I don't think that's what you meant, was it?

aegisknight on Mar 15, 2005

Wut? You're thinking too hard. It's a trivial answer in most languages.

def self(): return self

lordgalbalan on Mar 16, 2005

I see. I tried to read into your example that didn't seem to make much sense. ;) Seemed like you were trying for something grander.

Anyway, yeah sure you can always use a pointer to self to get its value. :)

anonymous on Mar 16, 2005

void f(){ return GetProcAddress(loadlibrary("this.dll"),"f"); }

anonymous on Mar 16, 2005

I don't think pointer comparison with == is type checked in C/C++.

But I could be wrong.

aegisknight on Mar 16, 2005

GetProcAddress doesn't return void. Like I said, you should actually test and run before you submit answers. ;)

sufianrhazi on Mar 16, 2005

Pointer comparison isn't necessarily typechecked, but that's not the problem. It's that function pointers must be aware of the type that they're returning. The problem is, you need to specify that functions need to return pointers of a function type which returns a pointer to a function type which returns... etc.

As for the casting, this is the best solution I came up with.

[HTML_REMOVED]

include <assert.h>

typedef void(fptr)();

fptr this(){ return this; }

int main(){ fptr one = this; fptr two = this(); fptr three = (fptr)two(); fptr four = (fptr)three(); assert(one == two); assert(two == three); assert(three == four); return 0; } [HTML_REMOVED]