I have the following question here
I need to understand the following MIPS disassembling code:
.text:00489060 la $v0, 0x4D0000
.text:00489064 la $v0, 0x4D0000
.text:00489068 addiu $v1, $v0, (aBig5 - 0x4D0000) # "BIG-5"
.text:0048906C lw $v0, (aBig5 - 0x4D0000)($v0) # "BIG-5"
.text:00489070 lbu $a0, (aBig5+5 - 0x4D6A68)($v1)
.text:00489074 lbu $v1, (aBig5+4 - 0x4D6A68)($v1)
.text:00489078 swl $v0, 0($s1)
.text:0048907C swr $v0, 3($s1)
.text:00489080 sb $a0, 5($s1)
.text:00489084 b loc_48A190
.text:00489088 sb $v1, 4($s1)
.text:0048908C # ---------------------------------------------------------------------------
.text:0048908C
.text:0048908C loc_48908C: # CODE XREF: cgiGetVar+B8j
.text:0048908C lw $v1, (dword_4D6A5C - 0x4D0000)($v0) #"ISO-"
.text:00489090 addiu $v0, (dword_4D6A5C - 0x4D0000)
.text:00489094 lw $a0, (dword_4D6A60 - 0x4D6A5C)($v0) #"8859"
.text:00489098 swl $v1, 0($s1)
.text:0048909C lbu $a1, (byte_4D6A66 - 0x4D6A5C)($v0) # zero byte
.text:004890A0 lbu $a2, (byte_4D6A64 - 0x4D6A5C)($v0) # "-"
.text:004890A4 lbu $v0, (byte_4D6A65 - 0x4D6A5C)($v0) # "1"
.text:004890A8 swr $v1, 3($s1)
.text:004890AC swl $a0, 4($s1)
.text:004890B0 swr $a0, 7($s1)
.text:004890B4 sb $a1, 0xA($s1)
.text:004890B8 sb $a2, 8($s1)
.text:004890BC b loc_48A190
.text:004890C0 sb $v0, 9($s1)
.text:004890C4 #
The task I need to solve is, to change Web Language ID from BIG-5 to ISO-8859-5 in my router firmware manager (without sources :) ) So ISO-8859-5 is 5 letters longer then BIG-5. I've moved the right hand (of BIG-5) text block in .rodata section )here is hex fragment:
4d6a50: 68746d6c 63686172 73657400 49534f2d htmlcharset.ISO-
4d6a60: 38383539 2d310000 4249472d 35000000 8859-1..BIG-5...
4d6a70: 53757043 484c616e 67000000 636f6e6e SupCHLang...conn
& changed the pointer in .text section. However when I run the program it loads just 6 bytes (I see "ISO-88" instead of (ISO-8859-5) in HTML-code page of router. So I guess I have to find where is the number of developing bytes is fixed and change it. Could somebody comments this code. Thanks a lot :)
The number of bytes is not specified. It just loads a word and two bytes and that's it:
; load one word (4 bytes) into v0
lw $v0, (aBig5 - 0x4D0000)($v0)
; load one byte into a0
lbu $a0, (aBig5+5 - 0x4D6A68)($v1)
; load one byte into v1
lbu $v1, (aBig5+4 - 0x4D6A68)($v1)
; store left part of the word from v0
swl $v0, 0($s1)
; store right part of the word from v0
swr $v0, 3($s1)
; store byte from a0
sb $a0, 5($s1)
; store byte from v1
sb $v1, 4($s1)
So it's basically fixed for a 5-char string and a longer one won't work without patching the code. You might be able to patch it for a 7-char one (8 bytes) by loading words and replacing the two sb by swl and swr. Another option, if you are certain that s1 is aligned, is to load three words (12 bytes) and store the latter two using just sw instead of swl/swr pair.