Question

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; 

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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).

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.