What does operator MyClass*() mean?

Question

I have the following class definition:

struct MyClass { int id; operator MyClass* () { return this; } };

I'm confused what the operator MyClass* () line does in the code above. Any ideas?

@kinjia "what the operator MyClass* () line does in the code above" - Nobody knows this except the author of the code.:) — Vlad from Moscow
– Vlad from Moscow, Commented Nov 5, 2017 at 9:52

StoryTeller - Unslander Monica · Accepted Answer · 2017-11-05 09:24:38Z

It's a type conversion operator. It allows an object of type MyClass to be implicitly converted to a pointer, without requiring the address-of operator to be applied.

Here's a small example to illustrate:

void foo(MyClass *pm) { // Use pm } int main() { MyClass m; foo(m); // Calls foo with m converted to its address by the operator foo(&m); // Explicitly obtains the address of m }

As for why the conversion is defined, that's debatable. Frankly, I've never seen this in the wild, and I can't guess as to why it was defined.

Whoever wrote this code probably tried to emulate references :P

msc · Accepted Answer · 2017-11-05 09:55:54Z

It is a user-defined conversion which is allow implicit or explicit conversion from class type to another type.

cppreference reference:

Syntax :

Conversion function is declared like a non-static member function or member function template with no parameters, no explicit return type, and with the name of the form:
operator conversion-type-id (1) explicit operator conversion-type-id (2) (since C++11) 
Declares a user-defined conversion function that participates in all implicit and explicit conversions

Declares a user-defined conversion function that participates in direct-initialization and explicit conversions only.

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