Re: initializing a member pointer

From:
Cupu <stefan.constantin@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 28 Nov 2007 03:56:21 -0800 (PST)
Message-ID:
<e621d318-4c6d-4eaa-a140-dbd59e44cb83@d27g2000prf.googlegroups.com>
On Nov 28, 1:44 pm, maj...@majsta.net wrote:

hello, I have probably a stupid question, but I don't get one thing.
In the following source, the first constructor "does not work" - it
creates a pointer variable pI, sets ii as *pI, but this created pI is
not the member one, the *C::pI is still uninitialised. In the later
constructor it's ok.
What's wrong with the first one?
Thank you for your help and time

m

#include <iostream>

class C
{
private:
        int *pI;
public:
        /**/
        // this constuctor does not work
        C(int ii) {
                int *pI = new int(ii);
        }
        /**/
        // this one is ok
        /*
        C(int ii) {
                int *pTmp = new int(ii);
                pI = pTmp;
        }
        */
        ~C() {delete pI;}
        void print()
        {
                std::cout << " *pI = " << *pI << std::endl;
        }};

        int main(int argc, char* argv[])
        {
                C d(3);
                d.print();
        }


Hi,

In the first constructor by doing "int *pI = new int(ii);" you
effectively declare a local int* variable named pI and hide the class
member C::pI, at scope end pI is destroyed and C::pI becomes visible
again.

If you don't redeclare pI (which is a mistake in this case) you'll get
the right behaviour since the compiler will translate that to this-

pI.


Cheers

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