Question

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Is there a difference in C++ between copy initialization and direct initialization?
Copy constructors and Assignment Operators

I have a class C in which I have overloaded Normal, copy constructor and assignment operator to print a trace of what is being called..

I wrote following pieces of code to test what is being called when?

C c1;                --> Normal Constuctor .. // understood Fine

C c2;
c2 = c1;             --> Normal constructor + assignment operator .. //understood Fine

C * c3 = new C(C1)   --> Copy constructor  // Understood Fine

C c4 = c1          --> copy constructor // Not Able to understand

This seems to baffle me since in this code though I am initializing at the time of declaration, it is through assignment operator and not copy constructor .. Am I understanding it wrong ??

Answer 1

Because C c4 = c1; is syntactically semantically equivalent to:

C c4(c1);

Both would invoke copy constructor in this particular case. However there is a subtle difference between “copy initialization” (1st syntax) and “direct initialization” (2nd syntax). Look at this answer.

_{Note: In “layman’s terms”, a variable (here c4) is constructed till the first ; (or ,‘ for multiple objects) is encountered; Till then everything is one or the other type of constructor.}

In case of C c4 = c1;, compiler doesn’t have to check for most vexing parse.
However one can disable C c4 = c1; kind of syntax by declaring copy constructor explicit. For that matter any constructor can be made explicit and you can prevent = sign in the construction.