Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
346 views
in Technique[技术] by (71.8m points)

c++ - Why is the assignment operator not called in this case in favor of the copy constructor?

From the wikipedia page for copy constructors:

X a = X();     

// valid given X(const X& copy_from_me) but not valid given X(X& copy_from_me)
// because the second wants a non-const X&
// to create a, the compiler first creates a temporary by invoking the default constructor
// of X, then uses the copy constructor to initialize as a copy of that temporary. 
// For some compilers both versions actually work but this behaviour should not be relied 
// upon because it's non-standard.

Specifically the part:

" the compiler first creates a temporary by invoking the default constructor of X, then uses the copy constructor to initialize as a copy of that temporary. "

My question is (assuming this is correct) why is this so? From the code, I would guess that the compiler would use the assignment operator after constructing an X.

I'm guessing its because assignment takes place in the same expression as initialization?

Also, what would be the reason to use this formula, rather than just a normal initialization X a; or if you want to copy X a(b); ?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Because the code is constructing an object. The = sign here is initializing, not assigning. You can only assign to an existing object, not to one that's under construction.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...