Wobien írta:
> &ulRandomNum is the adress of ulRandomNum (You don't give us the
> definition of it, so don't know what it is)
> (char *) is a cast. It makes the adress of ulRandomNum be read as a
> char array. So ((char *) &ulRandomNum) [nIndex]
> will give you a byte of the binary code of ulRandomNum.
> Wobien
>
> ----- Original Message ----- From: "Szentpali Janos" <ia...@ea...>
> To: <dev...@li...>
> Sent: Monday, January 07, 2008 6:32 PM
> Subject: [Dev-C++] Translation
>
>
> Hi,
>
> I am very new to C++. I only did VB programming... mostly. I am having
> some trouble with some C++ code. Could someone help me understand what
> the following means:
>
> --- start ---
>
> typedef DWORD(*FUNCPTR)(BYTE);
>
> DWORD ulWorkRandomNum;
> char *arr;
> int nIndex;
>
> ...
>
> arr [2 * nIndex] = 'A' + (((char *) &ulRandomNum) [nIndex] & 0x0f);
>
> --- end ---
>
> I am having trouble with "translating" (understanding) the
>
> ((char *) &ulRandomNum) [nIndex]
>
> part. I mean [nIndex] referes to the elemnt of which array :O
>
Thanks very much!
I knew about the cast and about the address operator, but it somehow
seamed strange to cut a pointer in pieces :O.
I thought that was the case, but I wasn't sure and I needed assurance
:D. Anyway the variable who's address you were talking about is just a
DWORD:
DWORD ulRandomNum;
Sorry I forgot to put it in the sample.
--
Szentpáli János
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