RE: [Dev-C++] C++ pointers

I saw that i made a mistake, so i resend.



> -----Original Message-----
> From: dev...@li...
> [mailto:dev...@li...] On Behalf 
> Of Green, Aaron
> Sent: Saturday, December 29, 2001 3:19 AM
> To: dev-c++ user's mailing list
> Subject: [Dev-C++] C++ pointers
> 
> 
> This may be OT, but I know I've seen programming questions
> here before, so I'll give it a try.
> 
> I wrote up a small program that tries to explain how pointers
> work to me.
> 
> #include <iostream>
> 
> double* f(double *pD2)
> {
> 	*pD2 = 15;
> 	return pD2;
> }
> 
> int main()
> {
> 	double d;
> 	double *pD = &d;
> 	d = 20;
> 	*pD = 10;
> 	std::cout << *pD << '\n';
> 	std::cout << d << '\n';

> 	double *pD3 = f(&d);


This points pD3 toy d which after the execution of this line is 15.
        

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	d = 100;


Now you change d toy 100 so *pD3 is now 100 (pD3==&d).

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	std::cout << *pD << '\n';
> 
> 
> 	return 0;
> }
> 
> 
> The last parts of the output are 100   100   100.  My 
> question is:  Why does *pD3 change when it points to a
> function that changes whatever is at &d ?  I thought calling 
> cout would run that function again, or at least give me the 
> same value as before.  


Calling cout<<*pD3; doesn't execute any function it simply the value of
d which is 100. If you wanted the function to be called again you should
do:

	std::cout << *(pD3=f(pD3)) << '\n';

    (or 	std::cout << *(pD3=f(&d)) << '\n';   )


or std::cout << *f(pD3) << '\n'

(    or  std::cout << *f(&d) << '\n'    )



Ioannis

* Ioannis Vranos
* Programming pages: http://www.noicys.f2s.com
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