How to convert a code in Assembly language to C code?

Question

i have assembly code and want to convert to equivalent in c i think it is
addition but i want if it is only simple addition so need help and thnx in advance :

Flash_Check_SUM:

 clr ZH
 clr ZL
 clr XH
 clr XL 

FCK_1: 

lpm R1,z+
add XL,R1 
brcc FCK_2
inc XH   ;carry set so inc Hi Bit

FCK_2:

cpi ZH,HIGH(BSTART*2 -16) 
brne FCK_1 
cpi ZL,LOW(BSTART*2 -16) 
brne FCK_1

lpm R0,Z+ 
lpm R1,Z 
cp R0,XH 
brne CK_Fail 
cp R1,XL 
brne CK_Fail 
ret
CK_Fail:

Answer 1

It would be something like this:

typedef union {
    unsigned short u16;
    unsigned char u8[2];
} Pointer;

const Pointer BSTART = ???;

int Flash_Check_SUM( Pointer Z, Pointer X ){

    unsigned char XH = ((unsigned char *)&X)[0];
    unsigned char XL = ((unsigned char *)&X)[1];
    unsigned char ZH = ((unsigned char *)&Z)[0];
    unsigned char ZL = ((unsigned char *)&Z)[1];

AddZ:
    unsigned short R1 = *Z++;
    X += *R1;
    if( XH > 0 ) goto IsEqual;
    XH++;

IsEqual:
    if( ZH != ((unsigned char *)&(BSTART * 2 - 16))[0] ) goto AddZ;
    if( ZL != ((unsigned char *)&(BSTART * 2 - 16))[1] ) goto AddZ;
    unsigned short R0 = *Z++;
    R1 = *Z;
    if( R0 != XH ) return -1;
    if( R1 != XL ) return -1;
    return 0;
}

Answer 2

Converting this to C code doesn't answer your actual question. A description of what the assembly is doing would be more helpful.

The X and Z registers are actually two registers each, so they are two bytes long. Some instructions use the Z register as an address to the input. That is what is happening in this code.

The function starts at address 0 and adds byte by byte to the 16-bit X register with

lpm R1,z+  ; load the byte at address Z into R1 and post-increment Z
add XL,R1  ; Add R1 to X

There is then some code to increment the high byte of X if the low byte overflows, (the usual addition algorithm).

The loop continues until Z reaches BSTART*2 -16, which appears to be all the bytes before the checksum.

The last part loads the stored checksum into R0 and R1, and then compares it to the checksum the code just computed in X (XH and XL).

If the checksum fails, then the code skips over the return statement and continues with the code following the checksum function, which presumably indicates an error condition of some sort to the user.