BinToDec:
mov eax, msg8
call sprint
mov edx, 255
mov ecx, sinput8
mov ebx, 0
mov eax, 3
int 80h
mov eax,sinput8
call atoiToDec
call iprintLFToDec
ret
;------------------------------------------
; void iprint(Integer number)
; Integer printing function (itoa)
iprintToDec:
push eax ; preserve eax on the stack to be restored after function runs
push ecx ; preserve ecx on the stack to be restored after function runs
push edx ; preserve edx on the stack to be restored after function runs
push esi ; preserve esi on the stack to be restored after function runs
mov ecx, 0 ; counter of how many bytes we need to print in the end
divideLoopToDec:
inc ecx ; count each byte to print - number of characters
mov edx, 0 ; empty edx
mov esi, 10 ; mov 10 into esi
idiv esi ; divide eax by esi
add edx, 48 ; convert edx to it's ascii representation - edx holds the remainder after a divide instruction
push edx ; push edx (string representation of an intger) onto the stack
cmp eax, 0 ; can the integer be divided anymore?
jnz divideLoopToDec ; jump if not zero to the label divideLoop
printLoopToDec:
dec ecx ; count down each byte that we put on the stack
mov eax, esp ; mov the stack pointer into eax for printing
call sprint ; call our string print function
pop eax ; remove last character from the stack to move esp forward
cmp ecx, 0 ; have we printed all bytes we pushed onto the stack?
jnz printLoopToDec ; jump is not zero to the label printLoop
pop esi ; restore esi from the value we pushed onto the stack at the start
pop edx ; restore edx from the value we pushed onto the stack at the start
pop ecx ; restore ecx from the value we pushed onto the stack at the start
pop eax ; restore eax from the value we pushed onto the stack at the start
ret
;------------------------------------------
; void iprintLF(Integer number)
; Integer printing function with linefeed (itoa)
iprintLFToDec:
call iprintToDec ; call our integer printing function
push eax ; push eax onto the stack to preserve it while we use the eax register in this function
mov eax, 0Ah ; move 0Ah into eax - 0Ah is the ascii character for a linefeed
push eax ; push the linefeed onto the stack so we can get the address
mov eax, esp ; move the address of the current stack pointer into eax for sprint
call sprint ; call our sprint function
pop eax ; remove our linefeed character from the stack
pop eax ; restore the original value of eax before our function was called
ret
;------------------------------------------
; int atoi(Integer number)
; Ascii to integer function (atoi)
atoiToDec:
push ebx ; preserve ebx on the stack to be restored after function runs
push ecx ; preserve ecx on the stack to be restored after function runs
push edx ; preserve edx on the stack to be restored after function runs
push esi ; preserve esi on the stack to be restored after function runs
mov esi, eax ; move pointer in eax into esi (our number to convert)
mov eax, 0 ; initialise eax with decimal value 0
mov ecx, 0 ; initialise ecx with decimal value 0
.multiplyLoopToDec:
xor ebx, ebx ; resets both lower and uppper bytes of ebx to be 0
mov bl, [esi+ecx] ; move a single byte into ebx register's lower half
cmp bl, 48 ; compare ebx register's lower half value against ascii value 48 (char value 0)
jl .finished ; jump if less than to label finished
cmp bl, 49 ; compare ebx register's lower half value against ascii value 49 (char value 1)
jg .finished ; jump if greater than to label finished
cmp bl, 10 ; compare ebx register's lower half value against ascii value 10 (linefeed character)
je .finished ; jump if equal to label finished
cmp bl, 0 ; compare ebx register's lower half value against decimal value 0 (end of string)
jz .finished ; jump if zero to label finished
sub bl, 48 ; convert ebx register's lower half to decimal representation of ascii value
add eax, ebx ; add ebx to our interger value in eax
mov ebx, 10 ; move decimal value 10 into ebx
mul ebx ; multiply eax by ebx to get place value
inc ecx ; increment ecx (our counter register)
jmp .multiplyLoopToDec ; continue multiply loop
.BinNum:
sub bl, 48
jmp BinLen
.finished:
mov ebx, 10 ; move decimal value 10 into ebx
div ebx ; divide eax by value in ebx (in this case 10)
pop esi ; restore esi from the value we pushed onto the stack at the start
pop edx ; restore edx from the value we pushed onto the stack at the start
pop ecx ; restore ecx from the value we pushed onto the stack at the start
pop ebx ; restore ebx from the value we pushed onto the stack at the start
ret
BinLen:
mov bl, [slen]
mul bl
mov bl, 2
我想使用nasm汇编语言将二进制转换为十进制。这只是我的代码的一部分，我面临的问题是当我输入我的输入（二进制）1101时，它只给我输出1101与我的输入相同。