Re: copy constructor not being called

On 15 abr, 20:33, suresh <suresh.amritap...@[EMAIL PROTECTED]
> wrote:
> Hi
>
> In the following code, I expect the copy constructor to be called as
> per the textbooks but it is not being called. Could you please tell me
> why? g++ version 4.1.2
>
> # include <iostream>
> using namespace std;
> class base
> {
>    char s;
> public:
>    base( ){cout<< "construction" << endl;}
>    base(const char a ) {cout << "construction with arg" << endl;}
>    base(const base & a){cout <<"copy constructor" << endl;}
>    base & operator=(const base & a){cout << "assignment operator" <<
> endl;}
>    ~base( ) {cout << "destruction" << endl;}
>
> };
>
> main ( )
> {
>    base b1 = 'x'; //why copy constructor not called here?
>    base b2 = b1; //copy constructor is called
>
> }
>
> the line base b1 = 'x', as per my understanding must create a
> tem****ary object and then it should be copied to b1. But only the
> constructor with argument is called in this example when executed.
>
> Could you please explain why?

Hello,

Indeed,

base b1 = 'x';

should construct a tem****ary object of type base using the constructor
of base that takes a const char argument, and then copy it into b1.
But the compiler is allowed to elide the call to the copy constructor
of base in order to generate more optimal code:

base b1('x');

Note that the compiler is still required to check whether the elided
copy constructor is accessible or not.

Regards,
David

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