Difference between func() and (*this).func() in C++

Question

I am working on someone else code in C++, and I found a weird call to a certain function func(). Here is an example:

if(condition)
    func();
else
    (*this).func();

What is the difference between func() and (*this).func()?

What are the cases where the call to func() and (*this).func() will execute different code?

In my case, func() is not a macro. It is a virtual function in the base class, with an implementation in both base and derived class, and no free func(). The if is located in a method in the base class.

Does the class have default values that are overwritten later? — Git, Aug 04 '16 at 10:12
Is this code located in the base class or in the derived class? — Sergey Kalinichenko, Aug 04 '16 at 10:24
Not that it matters much, but `(*this).func()` is more readable as `this->func()` — Déjà vu, Aug 04 '16 at 10:34
@pjc50 you cannot overload `*` for primitive types and `this` is a keyword which alvays evaluates to rvalue of pointer type. — Revolver_Ocelot, Aug 04 '16 at 11:49
@LukeG Debatable, certainly. But most of the code I see uses `->`, operator that has been created for a reason. Similarly, `a[X]` is usually preferred to `*(a+X)`. This makes me think that *"someone else code"* may not be doing what the author intended it to do ; maybe he thought the first one would call the base class implementation... — Déjà vu, Aug 04 '16 at 12:11
Do you want to provide the full context of the code so that people don't have to guess? — Barry, Aug 04 '16 at 16:50
Have you tried compiling to assembly and see what the compiler generates ? With g++, use -S — Vincent Fourmond, Aug 04 '16 at 20:46
@ringø I think LukeG misunderstood your first comment as "is more readable _than_", which is the opposite of what you meant...right? I understood it wrong at first glance, too. — DerManu, Aug 05 '16 at 01:39

score 49 · Accepted Answer · answered Aug 04 '16 at 10:16

49

There actually is a difference, but in a very non-trivial context. Consider this code:

void func ( )
{
    std::cout << "Free function" << std::endl;
}

template <typename Derived>
struct test : Derived
{
    void f ( )
    {
        func(); // 1
        this->func(); // 2
    }
};

struct derived
{
    void func ( )
    {
        std::cout << "Method" << std::endl;
    }
};

test<derived> t;

Now, if we call t.f(), the first line of test::f will invoke the free function func, while the second line will call derived::func.

answered Aug 04 '16 at 10:16

lisyarus

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2

I agree with you. But in my case, there are no free functions ;) – gtatr Aug 04 '16 at 10:18
40

@g.tataranni And, how should we know that, if there's no context in your question? Please provide us with [mcve], so that there's no guessing involved. – Algirdas Preidžius Aug 04 '16 at 10:22
11

I believe this was a joke about functions being free – slawekwin Aug 04 '16 at 10:47
25

This reminds me: know why the leaking program gets a rather large phone bill? No `free` calls. – Tobia Tesan Aug 04 '16 at 13:29
oh, stop it you ^^ – rain_ Aug 10 '16 at 06:50
1

So, this doesn't even answer the question, but gave me a silver C++ badge. Weird. – lisyarus Aug 10 '16 at 07:35

Galik · Answer 2 · 2016-08-04T14:56:04.580

31

It is impossible to tell from the snippet but possibly there are two callable objects called func(). The (*this).func(); makes sure the member function is called.

A callable object could be (for example) a functor or a lambda expression:

functor

struct func_type
{
    void operator()() const { /* do stuff */ }
};

func_type func; // called using func();

lambda

auto func = [](){ /* do stuff */ }; // called using func();

For example:

#include <iostream>

class A
{
public:

    // member 
    void func() { std::cout << "member function" << '\n'; }

    void other()
    {
        // lambda
        auto func = [](){ std::cout << "lambda function" << '\n'; };

        func(); // calls lambda

        (*this).func(); // calls member
    }
};

int main()
{
    A a;
    a.other();
}

Output:

lambda function
member function

edited Aug 04 '16 at 14:56

answered Aug 04 '16 at 10:16

Galik

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FTR, I had to overcome some inner resistance to upvote this answer and not downvote or edit away its misuse of the word [functor](https://en.wikipedia.org/wiki/Functor). Since _function object_ is a conventional standard meaning of the word _functor_ in the context of C++, what right have I to do that... still, just saying. I always cringe a bit when I see the word used this way. – leftaroundabout Aug 04 '16 at 19:34
@leftaroundabout I think many C++ programmers familiar with category theory (including me) feel the same internal rant. But that's the accepted and widespread terminology, we have to accept it. – lisyarus Aug 09 '16 at 19:54

score 17 · Answer 3 · answered Aug 04 '16 at 10:18

17

Another case when those two lines will call different functions:

#include <iostream>

namespace B
{ void foo() { std::cout << "namespace\n"; } }

struct A { 
  void foo() { std::cout << "member\n"; }

  void bar()
  {
      using B::foo;
      foo();
      (*this).foo();
  }
};

int main () 
{
    A a;
    a.bar();
}

answered Aug 04 '16 at 10:18

Revolver_Ocelot

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3
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48

There is no real difference to the other examples with one member and one non-member function, is there? – exilit Aug 04 '16 at 18:25
1

@exilit different paragraphs in lookup rules. All answers shows how different rules interact with each other. – Revolver_Ocelot Aug 04 '16 at 18:32

score 12 · Answer 4 · answered Aug 04 '16 at 10:17

12

With type dependent name, it may be different:

void func() { std::cout << "::func()\n"; }

struct S {
    void func() const { std::cout << "S::func()\n"; }
};

template <typename T>
struct C : T
{
    void foo() const {
        func();         // Call ::func
        (*this).func(); // Call S::func
    }
};

Demo

answered Aug 04 '16 at 10:17

Jarod42

203,559
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I've suddenly forgotten all my past CRTPs... Does this just do the same thing as `static_cast(this)->func()`? I just assumed the latter was strictly needed, as it's used in examples so often, but now I'm wondering why if `this->func()` works just as well. I suspect there's another reason I'm not seeing. – underscore_d Aug 05 '16 at 11:00
@underscore_d: The important part is *dependent name*. `func()` doesn't depend of `T` whereas `(*this).func()`, `this->func()` or `T::func()` does depend of T. – Jarod42 Aug 05 '16 at 11:07

Difference between func() and (*this).func() in C++

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