class A {};
class B { public: B (A a) {} };
A a;
B b=a;
Technically speaking, is a copy constructor being applied here on the creation of b ?
class A {}; class B { public: B (A a) {} }; A a;
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class A {};
class B { public: B (A a) {} };
A a;
B b=a;
Technically speaking, is a copy constructor being applied here on the creation of b ?
Yes…but probably not how you think.
A‘s copy constructor is being invoked on the creation of b, in order to do the pass-by-value of the parameterA aas a parameter to the B constructor.However, it is not running B’s copy constructor in the creation of b.EDIT: One learns something new every day. Apparently even more-technically-speaking, as @CharlesBailey pointed out…if you use the
B b = a;syntax (“copy initialization”) instead ofB b (a);syntax (“direct initialization”), a temporary value of type B might need to be created. At this point B’s copy constructor would wind up being called.It’s a little hard to study the phenomenon, but Charles points out that gcc has a
-fno-elide-constructorsoption (also: Wikipedia on Copy Elision) @JesseGood’s link has an exhaustive explanation and some demonstration code:Is there a difference in C++ between copy initialization and direct initialization?