I know the FooResult itself is constructed directly in the return value memory location. I suspect this is really just a case of typical move construction, but are there any guarantees about its members being moved?

struct FooResult {
    MoveConstructibleType x;
};

FooResult foo()
{
    MoveConstructibleType x;
    // do some stuff computing x...
    return FooResult {.x = x}; // is the construction of x guaranteed to be move constructed?
}

I know the FooResult itself is constructed directly in the return value memory location. I suspect this is really just a case of typical move construction, but are there any guarantees about its members being moved?

struct FooResult {
    MoveConstructibleType x;
};

FooResult foo()
{
    MoveConstructibleType x;
    // do some stuff computing x...
    return FooResult {.x = x}; // is the construction of x guaranteed to be move constructed?
}

• c++

• 1 the example would be more clear if not both variables were called x ;) – 463035818_is_not_an_ai Commented yesterday
• you want x to be moved from x ? – 463035818_is_not_an_ai Commented yesterday
• 3 NRVO is never guaranteed. and it applies to return var;, not return SomeObject{ var }; – Jarod42 Commented yesterday
• 2 x is copied here, you have to use return FooResult {.x = std::move(x)}; – Jarod42 Commented yesterday
• 1 cppinsights might help (NRVO variables are marked). – Jarod42 Commented yesterday

1 Answer 1

Actually, the key thing to remember here is that named variables aren't implicitly moved - so while it looks like a typical move construction case, the compiler won't automatically move x into FooResult unless std::move(x) is explicitly used. Otherwise, we fall back to copy semantics if a copy constructor exists.