On Jul 1, 11:49 am, Elias Salom?o Helou Neto <eshn...@gmail.com>
wrote:
On 1 jul, 00:24, Elias Salom?o 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 constr"
<< endl; }; ctest(int a) { cout <<"ctest int constor"
<<endl; };
ctest& operator= (ctest& a) { cout << "ctest copy
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 copy
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 copy
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() has
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
is _not_ const. If ctest had a non-const member function foo(), you could do
temporary. Your copy constructor takes its argument as a non-const
It doesn't. You also have to change the copy constructor.