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 If y.getTheVector() returns an object created in the function, x will point to an object that is no longer valid after the statement.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:
- The return value of
getTheVector()is a lvalue reference. In this case, the copy constructor is always invoked. - 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. - 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.
来源：https://stackoverflow.com/questions/23819358/does-this-copy-the-vector