malloc help (non-x86)

[chucara]

Hi, I'm new to assembler, and I'm trying to make a malloc function for the hitachi h/800 cpu.
It's standard AT&T syntax, with a few modifications.

I've read the hard- and software manual, as well as tutorials, but there's still a few things I don't grasp.

input:
start_adr = Start address in memory
byte         = byte value, which I want to overwrite the memory area with
size         = number of bytes to overwrite

Code Select Expand


malloc(int* start_adr, int byte, int size){
__asm__ __volatile (
    "add.w %0, %2"
    "0: cmp.w %0, %2"
       "beq 1f"
       "mov.b %1. @%0"
       "add.b %0, 1"
      "bra 0b"
     "1:"
    );
};

line 1 of the assembler code, I add the size og bytes to change to the starting address. Thus getting the end address.
line 2, I compare to see if the starting adress is equal to the end address (if it is, we are done)
line 3, I branch out of the loop, since the right number of bytes have been overwritten
line 4, I move the byte value to the start_adr location, which is the current byte to manipulate
line 5, I add one (1) to the address, to move to the next byte
line 6, I loop back to label 0

Now, the problem is about wheter to use word or byte (mov.b or mov.w). I want to move one byte at a time, so naturally it should be mov.b - but should it be high or low byte? And how do I get the next byte? Another cmp statement?

[Tedd]

If you're only ever pulling out one byte at a time, and only increasing the pointer by one each time, then there aren't any issue with byte-ordering.
You only have to worry about that if you read a word at a time, and then write a byte at a time.
However, if you read a word, then write a word, then add 2 to the pointer - it will still work fine; the internal read/write mechanism will sort out any byte ordering worries since they both use the same native format :wink

[James Ladd]

make it work a byte at a time, then later change it.
btw - why move bytes at all ? malloc does not usually clear memory.

[chucara]

Yeah, sorry about that.. The function is memset :D

And yes, it will do it one byte at a time.

[James Ladd]

Here is some example code for the x86. Hopefully this will help.
I did not write it, just modified it. Other here wrote it but I cant find that link right now.

Code Select Expand


align 4
memory_clear proc uses edi lpMem:DWORD, numBytes:DWORD, bUserVal:BYTE
    mov edi,lpMem       ; Pointer to memory
    mov  dl,bUserVal    ; Value to initialize memory with
    xor  eax,eax

    .if dl != 0         ; Build dword
        mov  al,dl
        shl  eax,8
        mov  al,dl
        mov  cx,ax
        shl  eax,16
        mov  ax,cx
    .endif

    mov  ecx,numBytes   ; number of bytes to write
    push ecx
    shr  ecx,2          ; numbers of dwords to write
    rep  stosd
    pop  ecx
    and  ecx,3          ; number of remaining bytes to write
    rep  stosb

    xor  eax,eax
    ret
memory_clear endp


This does some dwords at a time but then the remaining bytes are done.
You CAN do all bytes a byte at a time. Note: a BYTE at a time.