towi
towi

Reputation: 22347

C++0x: Capture By Value for Lambda, always a copy?

Is the compiler allowed to eliminate the copy that is required for the by-value capture?

vector<Image> movie1;
apply( [=movie1](){ return movie1.size(); } );

Upvotes: 14

Views: 1167

Answers (2)

Stack Overflow is garbage
Stack Overflow is garbage

Reputation: 248289

There is always the "as-if" rule. As long as it looks as if the rules had been followed, the compiler can do whatever it likes. So for objects where the copy constructor and destructor have no side effects, and where no changes are made to the copy, or the original object isn't accessed afterwards (so no one will notice if we make changes to the object), the compiler could prove that eliminating the copy is legal under the "as-if" rule.

But other than that, no, it can't just eliminate the copy, as @Ben said. The "regular" copy elision rules don't cover this case.

Upvotes: 3

Ben Voigt
Ben Voigt

Reputation: 283971

I'm fairly sure it cannot.

Even if the outer function no longer explicitly uses the variable, moving the variable would change the semantics of destruction.

Having move constructors for Image doesn't help, a vector can move or swap without moving its elements.

If the variable is read-only from this point forward, why not capture by reference? You could even create a const reference and capture that.

If the variable is not read-only, the copy is required. It doesn't matter whether the outer function or the lambda performs the modification, the compiler cannot allow that modification to become visible to the other.

The only difference I see between by-value capture and by-value argument passing is that the capture is named, it cannot be a temporary. So argument passing optimizations applicable to temporaries cannot be used.

Upvotes: 9

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