Re: Question

DarkD said:

> Im not sure you fully understand a pointer.

I'm not sure you do. :-)

> He is pointing to that memory
> location and displaying the integer value from that location. If you did
> this:
>
> int x;
> x = *&0x00c0;
> printf("Value at address 0x00c0 is %d\n",x);

You meant that to be:

int x;
x = *(int *)0x00c0;
printf("Value at address 0x00c0 is %d\n",x);


The code is, of course, non-portable, but may work anyway on some systems.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/2005
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
.

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