Home/ Questions/Q 87243
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-10T22:22:20+00:00

Given a class like this: class Foo { public: Foo(int); Foo(const Foo&); Foo& operator=(int);

  • 0

Given a class like this:

class Foo { public:     Foo(int);      Foo(const Foo&);      Foo& operator=(int);  private:     // ... };

Are these two lines exactly equivalent, or is there a subtle difference between them?

Foo f(42);  Foo f = 42;

Edit: I confused matters by making the Foo constructor ‘explicit’ in the original question. I’ve removed that, but appreciate the answers.

I’ve also added declaration of a copy constructor, to make it clear that copying may not be a trivial operation.

What I really want to know is, according to the C++ standard, will ‘Foo f = 42’ directly call the Foo(int) constructor, or is the copy constructor going to be called?

It looks like fasih.ahmed has the answer I was looking for (unless it’s wrong).

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Foo f = 42;

    This statement will make a temporary object for the value ’42’.

    Foo f(42);

    This statement will directly assign the value so one less function call.

    • 0