Is there no difference between a function address and the function address' address?

Question

void f()
{}

void test()
{
    auto fn_1 = f;
    auto fn_2 = &f;

    assert(fn_1 == fn_2); // OK

    fn_1();      // OK
    fn_2();      // OK
    (*fn_1)();   // OK
    (*fn_2)();   // OK
    (**fn_1)();  // OK
    (**fn_2)();  // OK
    (***fn_1)(); // OK
    (***fn_2)(); // OK
}

Are these behaviors explicitly defined by the C++ standard?

Check [this answer](http://stackoverflow.com/a/2795803/1229023). — raina77ow, Sep 18 '13 at 05:46
[Function calls](http://www.lysator.liu.se/c/rat/c3.html#3-3-2-2). — Dayal rai, Sep 18 '13 at 06:02

aaronman · Answer 1 · 2013-09-18T05:52:24.037

1

Yes the ampersand is optional, they produce the same result.

An lvalue of function type T can be converted to a prvalue of type “pointer to T.” The result is a pointer to the function.55

I'm just gonna go ahead and say if your using c++11 you should be using std::function anyway it's much easier to understand and use.

edited Sep 18 '13 at 05:52

answered Sep 18 '13 at 05:46

aaronman

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Could you refer me to the Holy standard? – xmllmx Sep 18 '13 at 05:48

score 1 · Answer 2 · answered Sep 18 '13 at 05:48

The issue at play here is that a function decays into a function pointer. The types of both of the variables fn_1 and fn_2 are void (*)(), i.e. "pointer to function taking no arguments and returning void". In the case of fn_1, the function f decays into a pointer to a function, while in the case of fn_2, you explicitly assign a pointer to a function to fn_2, and no decaying takes place.

Is there no difference between a function address and the function address' address?

2 Answers2