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Asked: May 10, 20262026-05-10T15:46:34+00:00 2026-05-10T15:46:34+00:00

I had assumed that the canonical form for operator+, assuming the existence of an

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I had assumed that the canonical form for operator+, assuming the existence of an overloaded operator+= member function, was like this:

const T operator+(const T& lhs, const T& rhs) {     return T(lhs) +=rhs; } 

But it was pointed out to me that this would also work:

const T operator+ (T lhs, const T& rhs) {     return lhs+=rhs; } 

In essence, this form transfers creation of the temporary from the body of the implementation to the function call.

It seems a little awkward to have different types for the two parameters, but is there anything wrong with the second form? Is there a reason to prefer one over the other?

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  1. 2026-05-10T15:46:35+00:00Added an answer on May 10, 2026 at 3:46 pm

    With the edited question, the first form would be preferred. The compiler will more likely optimize the return value (you could verify this by placing a breakpoint in the constructor for T). The first form also takes both parameters as const, which would be more desirable.

    Research on the topic of return value optimization, such as this link as a quick example: http://www.cs.cmu.edu/~gilpin/c++/performance.html

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