Why copy constructor is not called in this case?
c++, constructor, copy-constructor
Solution
From another comment: "So by default I should not rely on it (as it may depend on the compiler)"
No, it does not depend on the compiler, practically anyway. Any compiler worth a grain of sand won't waste time constructing an A, then copying it over.
In the standard it explicitly says that it is completely acceptable for `T = x;` to be equivalent to saying `T(x);`. (§12.8.15, pg. 211) Doing this with `T(T(x))` is obviously redundant, so it removes the inner `T`.
To get the desired behavior, you'd force the compiler to default construct the first A:
A a;
// A is now a fully constructed object,
// so it can't call constructors again:
a = A(5);
Problem
Here is the little code snippet: ``` class A { public: A(int value) : value_(value) { cout <<"Regular constructor" <<endl; } A(const A& other) : value_(other.value_) { cout <<"Copy constructor" <<endl; } private: int value_; }; int main() { A a = A(5); } ``` I assumed that output would be "Regular Constructor" (for RHS) followed by "Copy constructor" for LHS. So I avoided this style and always declared variable of class as `A a(5);`. But to my surprise in the code above copy constructor is never called (Visual C++ 2008) Does anybody know if this behavior is a result of compiler optimization, or some documented (and portable) feature of C++? Thanks.