When should I make explicit use of the `this` pointer?

Question

When should I explicitly write this->member in a method of a class?

Answer 1

Usually, you do not have to, this-> is implied.

Sometimes, there is a name ambiguity, where it can be used to disambiguate class members and local variables. However, here is a completely different case where this-> is explicitly required.

Consider the following code:

template<class T>
struct A {
   T i;
};

template<class T>
struct B : A<T> {
    T foo() {
        return this->i; //standard accepted by all compilers 
        //return i; //clang and gcc will fail
        //clang 13.1.6: use of undeclared identifier 'i'
        //gcc 11.3.0: 'i' was not declared in this scope
        //Microsoft C++ Compiler 2019 will accept it
    }

};

int main() {
    B<int> b;
    b.foo();
}

If you omit this->, some compilers do not know how to treat i. In order to tell it that i is indeed a member of A<T>, for any T, the this-> prefix is required.

Note: it is possible to still omit this-> prefix by using:

template<class T>
struct B : A<T> {
    int foo() {
        return A<T>::i; // explicitly refer to a variable in the base class 
        //where 'i' is now known to exist
    }

};

Answer 2

If you declare a local variable in a method with the same name as an existing member, you will have to use this->var to access the class member instead of the local variable.

#include <iostream>
using namespace std;
class A
{
    public:
        int a;

        void f() {
            a = 4;
            int a = 5;
            cout << a << endl;
            cout << this->a << endl;
        }
};

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

prints:

5
4

Answer 3

There are several reasons why you might need to use this pointer explicitly.

• When you want to pass a reference to your object to some function.
• When there is a locally declared object with the same name as the member object.
• When you're trying to access members of dependent base classes.
• Some people prefer the notation to visually disambiguate member accesses in their code.

Answer 4

Although I usually don't particular like it, I've seen others use this-> simply to get help from intellisense!

Answer 5

There are few cases where using this must be used, and there are others where using the this pointer is one way to solve a problem.

1) Alternatives Available: To resolve ambiguity between local variables and class members, as illustrated by @ASk.

2) No Alternative: To return a pointer or reference to this from a member function. This is frequently done (and should be done) when overloading operator+, operator-, operator=, etc:

class Foo
{
  Foo& operator=(const Foo& rhs)
  {
    return * this;
  }
};

Doing this permits an idiom known as "method chaining", where you perform several operations on an object in one line of code. Such as:

Student st;
st.SetAge (21).SetGender (male).SetClass ("C++ 101");

Some consider this consise, others consider it an abomination. Count me in the latter group.

3) No Alternative: To resolve names in dependant types. This comes up when using templates, as in this example:

#include <iostream>


template <typename Val>
class ValHolder
{
private:
  Val mVal;
public:
  ValHolder (const Val& val)
  :
    mVal (val)
  {
  }
  Val& GetVal() { return mVal; }
};

template <typename Val>
class ValProcessor
:
  public ValHolder <Val>
{
public:
  ValProcessor (const Val& val)
  :
    ValHolder <Val> (val)
  {
  }

  Val ComputeValue()
  {
//    int ret = 2 * GetVal();  // ERROR:  No member 'GetVal'
    int ret = 4 * this->GetVal();  // OK -- this tells compiler to examine dependant type (ValHolder)
    return ret;
  }
};

int main()
{
  ValProcessor <int> proc (42);
  const int val = proc.ComputeValue();
  std::cout << val << "\n";
}

4) Alternatives Available: As a part of coding style, to document which variables are member variables as opposed to local variables. I prefer a different naming scheme where member varibales can never have the same name as locals. Currently I'm using mName for members and name for locals.

Answer 6

1. Where a member variable would be hidden by a local variable
2. If you just want to make it explictly clear that you are calling an instance method/variable

Some coding standards use approach (2) as they claim it makes the code easier to read.

Example:
Assume MyClass has a member variable called 'count'

void MyClass::DoSomeStuff(void)
{
   int count = 0;

   .....
   count++;
   this->count = count;
}

Answer 7

The other uses for this (as I thought when I read the summary and half the question... .), disregarding (bad) naming disambiguation in other answers, are if you want to cast the current object, bind it in a function object or use it with a pointer-to-member.

Casts

void Foo::bar() {
    misc_nonconst_stuff();
    const Foo* const_this = this;
    const_this->bar(); // calls const version

    dynamic_cast<Bar*>(this)->bar(); // calls specific virtual function in case of multi-inheritance
} 

void Foo::bar() const {}

Binding

void Foo::baz() {
     for_each(m_stuff.begin(), m_stuff.end(),  bind(&Foo:framboozle, this, _1));        
     for_each(m_stuff.begin(), m_stuff.end(), [this](StuffUnit& s) { framboozle(s); });         
} 

void Foo::framboozle(StuffUnit& su) {}

std::vector<StuffUnit> m_stuff;

ptr-to-member

void Foo::boz() {
    bez(&Foo::bar);
    bez(&Foo::baz);
} 

void Foo::bez(void (Foo::*func_ptr)()) {
    for (int i=0; i<3; ++i) {
        (this->*func_ptr)();
    }
}

Hope it helps to show other uses of this than just this->member.

Answer 8

One other case is when invoking operators. E.g. instead of

bool Type::operator!=(const Type& rhs)
{
    return !operator==(rhs);
}

you can say

bool Type::operator!=(const Type& rhs)
{
    return !(*this == rhs);
}

Which might be more readable. Another example is the copy-and-swap:

Type& Type::operator=(const Type& rhs)
{
    Type temp(rhs);
    temp.swap(*this);
}

I don't know why it's not written swap(temp) but this seems to be common.

Answer 9

You only have to use this-> if you have a symbol with the same name in two potential namespaces. Take for example:

class A {
public:
   void setMyVar(int);
   void doStuff();

private:
   int myVar;
}

void A::setMyVar(int myVar)
{
  this->myVar = myVar;  // <- Interesting point in the code
}

void A::doStuff()
{
  int myVar = ::calculateSomething();
  this->myVar = myVar; // <- Interesting point in the code
}

At the interesting points in the code, referring to myVar will refer to the local (parameter or variable) myVar. In order to access the class member also called myVar, you need to explicitly use "this->".

Answer 10

The main (or I can say, the only) purpose of this pointer is that it points to the object used to invoke a member function.

Base on this purpose, we can have some cases that only using this pointer can solve the problem.

For example, we have to return the invoking object in a member function with argument is an same class object:

class human {

... 

human & human::compare(human & h){
    if (condition)
        return h;       // argument object
    else 
        return *this;   // invoking object
    }
};

Answer 11

You need to use this to disambiguate between a parameters/local variables and member variables.

class Foo
{
protected:
  int myX;

public:
  Foo(int myX)
  {
    this->myX = myX; 
  }
};

Answer 12

I found another interesting case of explicit usage of the "this" pointer in the Effective C++ book.

For example, say you have a const function like

  unsigned String::length() const

You don't want to calculate String's length for each call, hence you want to cache it doing something like

  unsigned String::length() const
  {
    if(!lengthInitialized)
    {
      length = strlen(data);
      lengthInitialized = 1;
    }
  }

But this won't compile - you are changing the object in a const function.

The trick to solve this requires casting this to a non-const this:

  String* const nonConstThis = (String* const) this;

Then, you'll be able to do in above

  nonConstThis->lengthInitialized = 1;