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

C++Dispatch

C++ Problem Overview


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.

C++ Solutions


Solution 1 - C++

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.

Solution 2 - C++

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

Solution 3 - C++

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();
}

Solution 4 - C++

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

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiongtatrView Question on Stackoverflow
Solution 1 - C++lisyarusView Answer on Stackoverflow
Solution 2 - C++GalikView Answer on Stackoverflow
Solution 3 - C++Revolver_OcelotView Answer on Stackoverflow
Solution 4 - C++Jarod42View Answer on Stackoverflow