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Why would a class/function have two overloads, one for lvalue and one for rvalue?
Eg, from this video, it says we have two overloads for vector<T>::push_back
void push_back( const T& value );
void push_back( T&& value );
Why can't we have just one overload by value,
void push_back( T value );
If it was an lvalue, value would be copied and if it was an rvalue, value would be moved. Isn't this the way how it works and guaranteed by the standard?
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You should definitely use emplace_back when you need its particular set of skills — for example, emplace_back is your only option when dealing with a deque<mutex> or other non-movable type — but push_back is the appropriate default. One reason is that emplace_back is more work for the compiler.
Specific use case for emplace_back : If you need to create a temporary object which will then be pushed into a container, use emplace_back instead of push_back . It will create the object in-place within the container. Notes: push_back in the above case will create a temporary object and move it into the container.
This code demonstrates that emplace_back calls the copy constructor of A for some reason to copy the first element. But if you leave copy constructor as deleted, it will use move constructor instead.
With your by-value proposition, technically there would be copy+move or move+move, whereas with the other two overloads there is a single copy or a single move.
80
Besides the point others mentioned, it would also require changing the old interface. And there are times when that's simply not acceptable.
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