C++ Instance Initialization Syntax

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 he'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.

Answer 2

class Foo {
public:
    Foo(explicit int);

    Foo& operator=(int);
};

That's invalid. The syntax is

class Foo {
public:
    explicit Foo(int);

    Foo& operator=(int);
};

The difference is that the conversion constructor cannot be used for implicit conversions when you put explicit before it:

Foo f(10); // works
Foo f = 10; // doesn't work

The above doesn't have anything to do with an assignment operator you declared there. It is not used since that is an initialization (only constructors are used). The following will use the assignment operator:

Foo f;
f = 10;

And will use the default constructor of Foo (the one taking no arguments).

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Edit: The questioner changed his question to the specific ways of whether

Foo f = 1; // called "copy initialization" 
Foo f(1);  // called "direct initialization"

Are the same. The answer is that they are equivalent to the following:

Foo f(Foo(1));
Foo f(1);

If and only if the conversion constructor taking the int is not declared with keyword explicit, otherwise the first is a compiler error (see above). The compiler is allowed to elide (optimize out) the temporary passed to the copy constructor of Foo in the first case if all semantic restrictions are still safisfied, and even if the copy constructor has side effects. That especially includes a visible copy constructor.