When should std::move be used on a function return value? [duplicate]

Answer 1

A C++ compiler is free to use std::move(foo):

• if it is known that foo is at the end of its lifetime, and
• the implicit use of std::move won't have any effect on the semantics of the C++ code other than the semantic effects allowed by the C++ specification.

It depends on the optimization capabilities of the C++ compiler whether it is able to compute which transformations from f(foo); foo.~Foo(); to f(std::move(foo)); foo.~Foo(); are profitable in terms of performance or in terms of memory consumption, while adhering to the C++ specification rules.

---

Conceptually speaking, year-2017 C++ compilers, such as GCC 6.3.0, are able to optimize this code:

Foo meh() {
    Foo foo(args);
    foo.method(xyz);
    bar();
    return foo;
}

into this code:

void meh(Foo *retval) {
   new (retval) Foo(arg);
   retval->method(xyz);
   bar();
}

which avoids calling the copy-constructor and the destructor of Foo.

---

Year-2017 C++ compilers, such as GCC 6.3.0, are unable to optimize these codes:

Foo meh_value() {
    Foo foo(args);
    Foo retval(foo);
    return retval;
}

Foo meh_pointer() {
    Foo *foo = get_foo();
    Foo retval(*foo);
    delete foo;
    return retval;
}

into these codes:

Foo meh_value() {
    Foo foo(args);
    Foo retval(std::move(foo));
    return retval;
}

Foo meh_pointer() {
    Foo *foo = get_foo();
    Foo retval(std::move(*foo));
    delete foo;
    return retval;
}

which means that a year-2017 programmer needs to specify such optimizations explicitly.