Thread: More help with pointers

I would like to transfer the byte in the memory chunk pData that is i bytes from the start of pData to the memory chunk newData that is (i+offset) bytes from the start of newData:

Is there a better way to code this:

Code:

	MemPtr	newData;
	Char	*pData
	Char	*offsetData;
	offsetData = (Char *)newData;
	offsetData += (i+offset);
	*offsetData = *pData+i;

this way:

Code:

	MemPtr	newData;
	Char	*pData
	((Char *)newData)[i+offset] = pData[i];

Yes, the second method loooks sane. Alternatively:

Code:

	MemPtr	newData;
        Char    *pNewData = (Char *)newData;
	Char	*pData
	pNewData[i+offset] = pData[i];

Just looks a bit tidier, probably makes very little difference in performance.

--
Mats

Originally Posted by MacGyver

Why the cast?

Code:

newData[i+offset] = pData[i];

Edit: This assumes Char is the same as char (ie. typedef or at least 1 byte in length) or was a typo.

I suspect MemPtr may be

Code:

typedef void *MemPtr;

That would necessitate a cast to be able to set it to something.

--
Mats

Mats,

You are correct about MemPtr.

Thank you for pointing me in the right direction.


MacGyver,

Char is char, Palm OS uses different names for reasons unknown to me.


Thanks

Greg

Originally Posted by gmrobert

Char is char, Palm OS uses different names for reasons unknown to me.


Thanks

Greg

It allows the OS to have control over the type used, without relying on what the compiler may do with things - and if someone decided to make everything for example Unicode (16-bit chars), then Char could just be redefined once, and all else would essentially work straight away [probably some breakages, but still easier than searching for ALL "char" in the code and replacing them with something else].

This is definitely very common in most OS's

--
Mats