--- Annamalai Gurusami <[EMAIL PROTECTED]>
wrote:
> Ray Devore wrote:
> >
> > > > Samp obj = Samp(12,22) // Explicit Call
> > Calls constructor (not copy constructor) for
> > Samp(12,22) and creates a temp object, then calls
> copy
> > constructor to create and initialize obj with the
> temp
> > object, then calls the destructor to delete the
> temp
> > object so you get two additional function calls
> with
> > the second method.
>
> I wrote a sample program to check if the copy
> constructor would be
> called for the above statement. But no copy
> constructor gets called.
> So I think there is no difference between the
> following two statements.
>
> Type obj(p);
> Type obj = p;
>
> Here is the sample program.
>
> (begin-code)
>
> #include <iostream>
> #include <string>
>
> using namespace std;
>
> class T
> {
> public:
> T(const string& n): name(n) {
> cout << "ctor: " << name << endl;
> }
> T(const T& that) {
> name = that.name;
> cout << "copy ctor: " << name << endl;
> }
> ~T() {
> cout << "dtor: " << name << endl;
> }
> private:
> string name;
> };
>
> int main()
> {
> T x("1");
> T y = T("2");
> }
>
> (end-code)
>
I added:
const T& operator=(const T& that)
{
name = that.name;
cout << "operator=: " << name << endl;
return *this;
}
and the assignment operator is not called either.
It appears that the compiler treats:
T y = T("2");
as if it was:
T y("2");
Thanks for the clarification.
Ray
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