How does `Foo obj = Foo(1)` work

Asked Jun 13 '14 at 07:34

Active Jun 13 '14 at 07:43

Viewed 146 times

Referring to the thread here: C++ object creation and constructor, the sentence Foo obj = Foo(1); should involve two object creations: one invoking Foo(int), the other Foo(const Foo&), namely copy-ctor.

class Foo{
    Foo(const Foo&){ //print "Foo(const Foo&) called" }
    Foo(int i){//print "Foo(int) called" }
    ...
}

...
Foo obj = Foo(1);

The above code only prints "Foo(int) called". How does it work in this case? Since I add printing functions in the ctors, it shouldn't be optimized out by the compiler, right?

edited May 23 '17 at 11:57

Community

asked Jun 13 '14 at 07:34

Neo1989

5

see about [Copy elision](http://en.wikipedia.org/wiki/Copy_elision) – jsantander Jun 13 '14 at 07:35
There is special permission to elide copies, even if it changes observable behaviour. – BoBTFish Jun 13 '14 at 07:36
@jsantander thanks for the pointers. This term "copy elision" is new to me. – Neo1989 Jun 13 '14 at 07:54

How does `Foo obj = Foo(1)` work

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