CODEWITHSUNDEEP
c++c++11move-semanticsmove-constructor
RippeR
RippeR

Reputation: 1472

Move semantics when sending object as function's parameter

I'm playing with move constructors and move assignments and i've stumbled on this problem. First code:

#include <iostream>
#include <utility>

class Foo {
    public:
        Foo() {}

        Foo(Foo&& other) {
            value = std::move(other.value);
            other.value = 1; //since it's int!
        }

        int value;

    private:
        Foo(const Foo& other);
};


void Bar(Foo&& x) {
    std::cout << "# " << x.value << std::endl;
}

int main() {
    Foo foo;
    foo.value = 5;

    Bar(std::move(foo));
    std::cout << foo.value << std::endl;

    return 0;
}

To my mind, when i use:

Bar(std::move(foo));

Program should "move" foo object to temp object created using move constructor in Bar function. Doing so would leave foo object's value equal zero. Unfortunatly it seams that object held in Bar function as parameter is some sort of reference, since it doesnt "move" original value but using Bar's parameter i can change it.

Would someone mind expalining me why i see in console:

#5
5

instead of

#5
0 //this should be done by move constructor?

Upvotes: 11

Views: 34325

Answers (4)

fider
fider

Reputation: 2036

Bar function should not take reference && (this syntax is valid only in move constructor declaration/definition):

void Bar(Foo x) { ... }  // not "Foo&& x"

To call move constructor of temporary argument object in function Bar:

Bar(  std::move( x )  );

To call copy constructor:

Bar( x );

Upvotes: 2

davidhigh
davidhigh

Reputation: 15468

Compare these two functions:

void Bar(Foo&& x) {
    std::cout << "# " << x.value << std::endl;
}

vs.

void Bar(Foo&& x) {
    std::cout << "# " << x.value << std::endl;
    Foo y=std::move(x);
}

Both take an rvalue reference, but only the second calls the move constructor. Consequently, the output of the first is

# 5
5

whereas the output of the second -- since the value of foo is changed -- is:

# 5
1

DEMO

EDIT: This is a question I had as well some time ago. My mistake then was to assume that the creation of an rvalue reference directly invokes the call of a move constructor. But, as was mentioned here before, std::move doesn't do anything at runtime, it just changes the type to an rvalue-reference. The move constructor is only invoked when you "move" your rvalue reference into another object as above.

Upvotes: 2

vsoftco
vsoftco

Reputation: 56547

When you write value = std::move(other.value); you should understand that std::move does not move anything. It just converts its parameter to a rvalue reference, then, if the left hand side has a move constructor/assignment operator, the left hand side deals with it (and how to actually move the object). For plain old types (PODs), std::move doesn't really do anything, so the old value remains the same. You are not "physically" moving anything.

Upvotes: 9

sehe
sehe

Reputation: 392921

An rvalue reference is (surprise:) a reference indeed.

You can move from it, but std::move does not move.

So, if you don't move from it, you'll actually operate on the rvalue object (through the rvalue reference).

The usual pattern would be

void foo(X&& x)
{
    X mine(std::move(x));

    // x will not be affected anymore
}

However, when you do

void foo(X&& x)
{
    x.stuff();
    x.setBooDitty(42);
}

effectively, X&& is just acting as a traditional reference

Upvotes: 25

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