In the programming realm, the usage of std::move on a function's return value prompts the question of when such a move is warranted. Consider the following code snippet:
struct Foo {}; Foo meh() { return std::move(Foo()); }
In this case, the move seems redundant since the newly created Foo is already an xvalue due to its temporary nature. However, in scenarios like the one below, std::move plays a crucial role:
struct Foo {}; Foo meh() { Foo foo; // Perform operations, ensuring safe disposal of foo return std::move(foo); }
Here, foo is created as an lvalue, and without std::move, the compiler might not discern that it can be safely removed afterward. The move operation explicitly requests that the object be moved rather than copied, ensuring proper resource management.
A key concept to grasp is the "as-if" rule outlined in the C standard (12.8/32). This rule mandates that when specific conditions are met for eliding a copy operation, the compiler must behave as if the object were designated by an rvalue, even if it is an lvalue. In the given example, NRVO (Named Return Value Optimization) allows the copy from foo to the return value to be elided. Consequently, the move constructor is automatically invoked without the explicit std::move.
In summary, std::move should be used on a return value when the following criteria are met:
To simplify the decision-making process in non-template code, consider these rules:
By adhering to these simplified rules, developers sacrifice some move elision but gain the advantage of ease of use. However, for complex templates or types with expensive move operations, meticulous analysis is recommended to ensure optimal code performance.
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