Does this copy the vector?

问题:

If I have the following code, is the vector copied?

std::vector<int> x = y.getTheVector();

or would it depend on whether the return type of getTheVector() is by reference?

or would I just need to use:

std::vector<int>& x = y.getTheVector();

or, would I need to do both?

回答1:

std::vector<int> x = y.getTheVector();

always makes a copy, regardless of the return type of y.getTheVector();.

std::vector<int>& x = y.getTheVector();

would not make a copy. However, x will be valid as long as y.getTheVector() returns a reference to an object that is going to be valid after the function returns. If y.getTheVector() returns an object created in the function, x will point to an object that is no longer valid after the statement.

回答2:

std::vector<int> x = y.getTheVector();

This is copy-initialization. There are three possible scenarios:

1. The return value of getTheVector() is a lvalue reference. In this case, the copy constructor is always invoked.
2. The return value of getTheVector() is a temporary. In this case, the move constructor may be called, or the move/copy may be completely elided by the compiler.
3. The return value is a rvalue reference (usually a terrible idea). In this case, the move constructor is called.

For this line,

 std::vector<int>& x = y.getTheVector();

This only compiles if getTheVector returns a lvalue reference; a temporary cannot be bound a non-const lvalue reference. In this case, no copy is ever made; but the lifetime problem may be tricky.

回答3:

std::vector<int> x = y.getTheVector();

Your first example does indeed copy the vector, regardless of whether the "getTheVector" function returns a vector or a reference to a vector.

std::vector<int>& x = y.getTheVector();

In your second example, however, you are creating a reference, so the vector will NOT be copied.