I have two copy constructors
Foo(Foo &o开发者_开发百科bj){
}
Foo(Foo *obj){
}
When will the second copy constructor will get called?
No you don't - you have one copy constructor
Foo( Foo * obj );
is not a copy constructor, and will never be used as such by the C++ compiler. You can of course use it yourself:
Foo a;
Foo b( & a ); // use your constructor
Note also that your real copy constructor should be declared as;
Foo( const Foo & f );
though the lack of const does not prevent it from being a copy constructor.
Leaving aside that the second constructor isn't a copy constructor - you actually wanted to know when the second constructor will be called.
The Foo(Foo* obj);
constructor is a single parameter constructor - because it hasn't been marked with the explicit
keyword, it provides an implicit conversion from Foo*
to Foo
. It may be called at any time where a Foo*
is used in the place of a Foo
or const Foo&
- if it's being called unexpectedly, that's almost certainly what's happening.
In general, single parameter constructors should either be copy constructors (which other answers have explained) or should be marked explicit
. Constructors which provide implicit conversions should be used sparingly.
The second is not a copy constructor. It is a constructor that gets called when you create a new Foo object, giving a pointer to a Foo as parameter.
Foo foo0;
Foo foo1 = foo0; // Calls copy constructor
Foo foo2(foo0); // Calls copy constructor
Foo foo3(&foo0); // Calls constructor taking a pointer as parameter
One of your constructors is a copy constructor, the other is just a normal constructor.
The second will be called if you explicitly initialize a Foo
from a pointer to Foo
or in other situations where an conversion from a pointer to Foo
to an r-value Foo
is called for, such as argument passing and function returns.
It's usually a bad idea to have such an implicit conversion; it may occur when you don't expect it to and is liable to turn trivial typos from compile errors into unusual behaviour at runtime.
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