CODEWITHSUNDEEP
c++move-semantics
user3983738
user3983738

Reputation: 29

Is this valid optimization?

"Hello." is a temporary that's constructed, copied into std::string and subsequently destroyed. We can skip copying and go straight to moving. But any decent compiler will elide the copy. So is there any point to the following:

std::string s(std::move("Hello."));

Upvotes: 0

Views: 86

Answers (3)

Slava
Slava

Reputation: 44238

Your code does not make much sense as it would be logically equal to this:

Foo bar( std::move( 123 ) );

There is no point to move constant, and string literal is a constant. Better question would be, and I think is that what you really meant by your question:

class Foo { ... };

class Bar {
public:
    Bar( const Foo &foo );
    Bar( Foo &&foo );
...
};

Bar bar( std::move( Foo() ); // Does it make sense?

So actually for this case std::move can be used as a syntax sugar to avoid function declaration of the form:

Bar bar( Foo() );

But it would not make any optimization and in this case:

Bar bar = std::move( Foo() );

It would be completely redundant

Upvotes: 0

CashCow
CashCow

Reputation: 31435

If that's even legal and I don't know if it is, it is not optimising anything.

The literal is created at compile time in static space. It can't be moved. string will always own a copy of the data and not just "refer" to it.

Upvotes: 0

Mike Seymour
Mike Seymour

Reputation: 254431

No, there's no point moving a string literal. It's a static array (not a "temporary that's constructed"), which can't be moved, only copied.

Upvotes: 3

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