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rvalue references - Why aren't they forwarded?

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Consider this code:
[i]void f(int &&value)
{ std::cout << "Called: " << __PRETTY_FUNCTION__ << std::endl; }[/i]
[i]void f(const int &value)
{ std::cout << "Called: " << __PRETTY_FUNCTION__ << std::endl; }[/i]

[i]void g(int &&value)
{
std::cout << "Called: " << __PRETTY_FUNCTION__ << std::endl;
f(value);
}[/i]
[i]void g(const int &value)
{
std::cout << "Called: " << __PRETTY_FUNCTION__ << std::endl;
f(value);
}[/i]

[i]int main(int argc, char *argv[])
{
std::cout << "Temporary:" << std::endl;
g(123);

std::cout << "\nNon-temporary:" << std::endl;
int value = 123;
g(value);

return 0;
}[/i]


It outputs the following:
[i]Temporary:
Called: void g(int&&)
Called: void f([b][color=#ff0000]const int&[/color][/b]) [color=#008080]//I was expecting[b] f(int&&)[/b] to be called.[/color][/i]

[i]Non-temporary:[/i]

[i]Called: void g(const int&)
Called: void f(const int&)[/i]

Why isn't the rvalue reference passed to the f() overload taking an rvalue reference?

I can force it to with:
[i]void g(int &&value)
{
f(std::move(value));
}[/i]

...but I assumed it'd be propagated automatically.

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I just learned about rvalue references myself, so I am not sure, but the behavior you are seeing is what I expected. Within `g(int &&value)', the parameter `value' is not a temporary, so the compiler doesn't treat it as one.

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What sicrane said. The idea is that you need to explicitly ask for move semantics to avoid bad surprises. You don't want some function calls to steal your objects through move semantics without explcitly asking for it, which you do using std::move.

If you call a function or assign something and want to move the value instead of copying it, you do something like someFunction( move( thing ) ) or anotherthing = move( thing ). This way you know that thing is not valid anymore after that.

(btw I don't remember what the standard says about the state of an object after it has been moved, iirc it's just up to the move constructor or move operator to make sure the object is in a state where its destructor won't blow up once it goes out of scope) Edited by Zlodo

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[quote name='Zlodo' timestamp='1352324550' post='4998597']
(btw I don't remember what the standard says about the state of an object after it has been moved, iirc it's just up to the move constructor or move operator to make sure the object is in a state where its destructor won't blow up once it goes out of scope)
[/quote]
Nothing happens to the object itself when it is moved. It is not the object that is moved, but it is the object's resources that have changed owners. So it is up to the move constructors/operators to ensure that the resources are transferred properly and that the object on the right hand side is still valid, although empty since its resources have been transferred to another object. Edited by Brother Bob

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All the standard says is that the constructed/assigned object has the state of the moved object and the moved object has a valid state (which does not actually have to be empty). So for a move assignment operator, being the same as the copy assignment operator is valid, taking ownership of the moved object's resources and freeing the original's resources is valid, or being the same as a swap is also valid.

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void g(int &&value)
{
std::cout << "Called: " << __PRETTY_FUNCTION__ << std::endl;
f([b]std::forward[/b](value));
}

I never used perfect forward before, but I think you need explicitly cast value to a rvalue, like what std::forward does.

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