Re: class instantiation question
On Jul 1, 1:12 pm, Kai-Uwe Bux <jkherci...@gmx.net> wrote:
hill....@gmail.com wrote:
On Jul 1, 11:49 am, Elias Salom=E3o Helou Neto <eshn...@gmail.com>
wrote:
On 1 jul, 00:24, Elias Salom=E3o Helou Neto <eshn...@gmail.com> wrote:
On 1 jul, 00:16, "hill....@gmail.com" <hill....@gmail.com> wrote:
Hi,
I stuck into this problem that I can't figure it out.
Here is the class definition:
class ctest {
public:
ctest(void) { cout << "ctest default constor" <<
endl; }; ctest(ctest& c) { cout <<"ctest copy cons=
tr"
<< endl; }; ctest(int a) { cout <<"ctest int const=
or"
<<endl; };
ctest& operator= (ctest& a) { cout << "ctest cop=
y
assignment" <<
endl; return *this; };
operator int() { cout << "call operator int()"
<<endl; return 20; }
};
int main(void)
{
ctest cc = ctest();
}
And it outputs as following:
ctest default constor
call operator int()
ctest int constor
I wonder why it will invoke "operator int()" and call "constructor
with int as argument".
Thanks
Brad
I wonder where is the
return( 0 );
statement.
Well, it does not quite solve the problem. In fact, the issue here is
const-correctness, because ctest() must be used as const, but ctest
has ctest& or int as arguments, being the second the only possible the
compiler tryes to find a conversion. The following does not compile:
#include <iostream>
class ctest {
public:
ctest(void) { std::cout << "ctest default constor" <<
std::endl; };
ctest(ctest& c) { std::cout <<"ctest copy constr" <<
std::endl; };
//ctest(int a) { std::cout <<"ctest int constor" <<
std::endl; };
ctest& operator= (ctest& a) { std::cout << "ctest co=
py
assignment" << std::endl; return *this; };
//operator int() { std::cout << "call operator int()"
<< std::endl; return 20; }
};
int main(void)
{
ctest cc = ctest();
return( 0 );
}
but the following does:
#include <iostream>
class ctest {
public:
ctest(void) { std::cout << "ctest default constor" <<
std::endl; };
ctest(const ctest& c) { std::cout <<"ctest copy
constr" << std::endl; };
ctest(int a) { std::cout <<"ctest int constor" <<
std::endl; };
ctest& operator= (ctest& a) { std::cout << "ctest co=
py
assignment" << std::endl; return *this; };
operator int() { std::cout << "call operator int()" <<
std::endl; return 20; }
};
int main(void)
{
ctest cc = ctest();
return( 0 );
}
and prints
ctest default constor
only once, because of optimization, I guess.
Forget to mention, I use Microsoft C/C++ Optimizing Compiler Version
14.00.50727.42 without any optimization option.
I comment out ctest(int a) and operator int() it still compiles. And
the execution output is
"ctest default constor"
From your reply and my understanding right now, ctest cc = ctest() ha=
s
two part. First part is that ctest() will create a temp ctest object
using default constructor. And second part is to create object cc
using copy constructor with that temp ctest as argument. Because that
temp ctest is a const, and my original class definition has no const-
copy constructor; so compiler convert that temp ctest into int and use
ctest(int a) to create object cc.
That understanding is not entirely correct. The temporary
ctest()
is _not_ const. If ctest had a non-const member function foo(), you could=
do
ctest().foo()
and the compiler would not complain.
The misconception that temporaries are const stems from the provisions in
[8.5.3/5] that prohibit initializing a non-const reference from a
temporary. Your copy constructor takes its argument as a non-const
reference and since the compiler cannot use the temporary ctest() to
initialize it, it rejects this option.
But why it still compiles after comment out ctest(int a) and operator
int()?
It doesn't. You also have to change the copy constructor.
And, after I add const to copy constructor like this: ctest(const
ctest& c) { cout <<"ctest copy constr" << endl; }, why it doesn't
output "ctest copy constr"?
Because the compiler is allowed to optimize the copy constructor call awa=
y
even if that changes the observable behavior of the program.
Best
Kai-Uwe Bux
Now I know why my original class definition will produce those
outputs.:-)
But I refine class definition like Elias suggested:
class ctest {
public:
ctest(void) { std::cout << "ctest default constor" <<
std::endl; };
ctest(ctest& c) { std::cout <<"ctest copy constr" <<
std::endl; };
ctest& operator= (ctest& a) { std::cout << "ctest copy
assignment" << std::endl; return *this; };
};
I build this successfully under Visual Studio 2005 and "ctest cc =
ctest()" outputs "ctest default constor".
Why is that?