Why does a function declaration within another function compile and what does it do?

Question

I intended to call a private class member function, but by a copy&paste mistake pasted the line as this function is declared in the header file:

void DebugView::on_cbYAxisEnabled_stateChanged(int)
{
    void updateAxisEnabled();
}

instead of

void DebugView::on_cbYAxisEnabled_stateChanged(int)
{
    updateAxisEnabled();
}

Surprisingly, the code was compiled and executed. However the method updateAxisEnabled() was not executed.

So, why does it compile? Was here a local function declared within a method body or has void instructed the compiler to ignore whatever comes afterwards?

The compiler is Visual Studio 2008.

P.S.: I'm aware of class declaration/definition within functions, but not functions within functions in C++.

P.S.: I'm aware of class declaration/definition within functions, but not functions within functions in C++ Well, now you know that's possible too. — deviantfan
– deviantfan, Commented Oct 13, 2015 at 12:38
@deviantfan: I'm sure, that it is not legal to DEFINE a local function in C++. So what is the point to DELCARE a local function without being able to define a body for it? — Valentin H
– Valentin H, Commented Oct 13, 2015 at 12:41
@ValentinHeinitz Have you tried with g++ for example,I am curious how that would work. — Richard Rublev
– Richard Rublev, Commented Oct 13, 2015 at 12:43
The same reason you forward declare anything; telling the compiler that a symbol's definition exists elsewhere. — AndyG
– AndyG, Commented Oct 13, 2015 at 12:43
@RichardRublev: yes, seems to work, Tried here: gcc.godbolt.org — Valentin H
– Valentin H, Commented Oct 13, 2015 at 12:45

Peter Mortensen · Accepted Answer · 2015-10-13 21:52:24Z

38

void updateAxisEnabled(); is a function declaration.

Sample:

#include <cstdio>

void a();
void b();

int main(void) {
    a();
    b();
    return 0;
}

void a() {
    void c(); // Declaration
    c(); // Call it
}

void b() {
    c(); // Error: not declared
}

void c() {
    puts("Hello, world!");
}

edited Oct 13, 2015 at 21:52

Peter Mortensen

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answered Oct 13, 2015 at 12:32

MikeCAT

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1 Comment

Valentin H Over a year ago

Cool! It works. I've tried it here: gcc.godbolt.org. Good to know, that it's possible to make forward declaration in a member function. Thanks!

edmz · Accepted Answer · 2015-10-13 17:42:42Z

5

It is perfectly allowed to declare a function inside a function scope: a function may be declared in any scope.

A common mistake among C++ programmers is indeed to:

void foo()
{
    MyObject bar(); // 1
    bar.someMethod(); // 2
}

this will miserably fail to compile because line 1 is not declaring a MyObject named bar and calling its constructor explicitly; rather, it is declaring a function named bar that returns a MyObject. Thefore, there is really no object to call someMethod on.

answered Oct 13, 2015 at 17:42

edmz

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6 Comments

Justin Over a year ago

This is why I like using this syntax to initialize my variables: MyObject bar{};

edmz Over a year ago

@Just Yes, the uniform initialization can solve this usses since C++11.

Brian Bi Over a year ago

Note, however, that you can't declare a non-member function at class scope, unless you make it a friend.

Valentin H Over a year ago

@Brian: what is the syntax fo it?

Valentin H Over a year ago

@Brian Weird! One can even define a global not-member function inside class declaration scope. Seems to be another case for: mindprod.com/jgloss/unmain.html :-)

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