Problems with pointer arithmetic.

Question

5.00/5 (1 vote)

See more:

Hi,
Say I've allocated some memory using malloc(), the operating system will give me the address to the memory allocated (i.e. a page). Now I am trying to browse this area of memory as if it was a 16-bit address space which starts from 0 and ends at 0xFFFF. (Don't ask why!, Just wanted to understand some concepts)
So I wrote this routine which will convert the so-called "virtual" address into a physical one:

C

uint32_t* virtual_to_physical(uint16_t OFFSET, uint32_t* MEMORY_BUFFER)
{
	uint32_t* a = MEMORY_BUFFER + OFFSET;
	return a;
}

Basically this routine will get the malloc()ed buffer and then will add the offset and return the physical address.
Now when I try it like this:

C

uint32_t* physaddr = (uint32_t*)malloc(0xFFFF);
printf("\nmalloc()ed buffer: %p", physaddr);
uint32_t* new_addr = virtual_to_physical(1, physaddr); //! Get the physical address of offset 1 in the malloc()ed buffer
printf("\nVirtual Address 1 turned to physical address: %p", (uint32_t*)new_addr);

But the output I get is:

malloc()ed buffer: 0x12b90a0
Virtual Address 1 turned to physical address: 0x12b90a4

Instead of adding 1 to the Physical address it increments it by 4?

Posted 23-May-14 21:55pm

sid2x

Updated 23-May-14 21:57pm

v2

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2014-05-23T22:34:00

Yes. It does - that is what I would expect.
Think about it: What datatype is your memory buffer? It's a uint32_t (because MEMORY_BUFFER is declares as a pointer to uint32_t) which means it addresses 32 bit quantities, so when you add 1 to the pointer it moves it to the next uint32_t values, not the next byte in the memory. Physical addresses are always in bytes - or you wouldn't save any memory space when you used characters r byte values - but pointers always point to a "whole" value of the type of the pointer - or moving up through an array of integers would be a PITA! :laugh:

If you want to work in byte increments, then use a pointer to an unsigned char, or other 8 bit datatype in your system.