Operator -> for shared pointer

Question

Hi

I have a member variable T* m_ptr into templated class. This is in line with shared pointer implementation.

My query is operator overloaded function as below:

T* operator->()
{
return m_ptr;
}

Why this pointer directly call method ?

Should it not be mysharedptr->->mymethod()

Double time -> is not needed?

Answer 1

when "operator->" returns, the "operator->" is applied to the value returned, with the original second operand.

so :
"bptr->display(); "

translates into:
"(bptr.operator->())->display();"

chained calls of ".operator->()" can be reduced to a single "->" .

C++98 standard §13.5.6/1 "Class member access:
"
An expression x->m is interpreted as (x.operator->())->m for a class object x of type T if T::operator->() exists and if the operator is selected as the best match function by the overload resolution mechanism.
"

Answer 2

Thanks

Answer 3

-> was not available to use before you overloaded it. How can there be the first -> in '->->' ? If you can already use ->, why is there the need to define an operator overlaod in the first place?

Answer 4

MO ELomari
nice.

Answer 5

MO ELomari thanks

Answer 6

T* operator->() const is the concern here for me.

This is called from last line of the main function.

bptr->display() results into printing display by calling a function display.

Now comes tricky part.

bptr is object of class custom shared pointer. Overloaded -> operator I.e. bptr-> will gives us T* from class member. This T* should need now another -> to call display method.

In other word , Should it not be (bptr->)->display()?
https://sololearn.com/compiler-playground/chbZHAzzB0Rz/?ref=app