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我使用汇编语言（nasm）做了一个程序来进行重叠块传输，即如果我的数组中有一个数组包含'10,20,30,40,50'（不含引号）然后重叠例如2个元素后，我的结果数组应包含'10,20,10,20,30,40,50'（不含引号）。但我的问题是，当我显示我的结果数组时，它只显示'10,20,10,20,30'（不含引号）。我无法弄清楚这个问题。下面显示的是我的代码。任何帮助将不胜感激。无法在程序集中执行重叠块传输
%macro disp 2 ; Display macro
mov rax,1
mov rdi,1
mov rsi,%1
mov rdx,%2
syscall
%endm
%macro accept 2 ; Accept Data
mov rax,0
mov rdi,0
mov rsi,%1
mov rdx,%2
syscall
%endm
global _start
section .data ; Data Section
arr db 10h,20h,30h,40h,50h
msg1: db "",10,"Input array is",10
len1: equ $-msg1
msg2: db "",10,"Output array is",10
len2: equ $-msg2
msg3: db "Enter the number to be overlapped",10
len3: equ $-msg3
section .bss ; Bss Section
arr1 resb 10
ar1 resb 10
ar2 resb 10
cn resb 2
section .text ; Text Section
_start: ; Tell linker entry point
disp msg3,len3
accept cn,2 ; Accept No. of overlaps to be done from user
mov cl,[cn] ; Convert the cn from ASCII
sub cl,'0' ; to Decimal
mov rsi,arr
mov rdi,arr1
up: mov al,[rsi] ; Copy the contents from arr to arr1 for cn times
mov [rdi],al
inc rsi
inc rdi
dec cl
jnz up
mov rsi,arr ; Now copy the contents again from starting
mov cl,5
up_a: mov al,[rsi]
mov [rdi],al
inc rsi
inc rdi
dec cl
jnz up_a
mov rsi,arr ; Convert to ASCII
mov rdi,ar1
mov rdx,5
call asci
mov rsi,arr1 ; Convert to ASCII
mov rdi,ar2
mov rdx,10
call asci
disp msg1,len1
disp ar1,10
disp msg2,len2
disp ar2,10
mov rax,60
mov rdi,0
syscall
asci:
up1: mov al,[rsi] ;Move the first element pointed by rsi into al register
mov cl,2 ;The loop counter (there are 2 digits)
up2: rol al,4 ; Rotate the contents of al 4 bits to the left. What were
; previously the most significant bit will now be in
; the least significant bit of al. This is done because
; we want to print the most significant digit first.
mov bl,al ; Make a copy of the rotated version of al.
and al,0Fh ; Keep the least significant bit of al and set all other
; bits of al to 0.
cmp al,09h ; al will now be in the range 0..15. Is it greater than 9?
ja dn1 ; ..if so, jump to dn1.
add al,30h ; al was in the range 0..9. The characters '0'..'9' are encoded as
; 30h..39h in ASCII, so add 30h to convert into a character.
jmp dn2 ; We're done with this case.
dn1: add al,37h ; al is in the range 10..15. The characters 'A'..'F' are encoded as
; 41h..46h in ASCII, so add 37h to convert into a character.
dn2: mov [rdi],al ; Store the character in the buffer pointed to by rdi.
mov al,bl ; Restore al to the value it had right after the rol. So on the
; next iteration we'll be processing what were originally the
; second most significant bit, and so on.
inc rdi ; Increment the buffer pointer.
dec cl ; Decrement the loop counter.
jnz up2 ; Repeat for all 2 digits.
inc rsi ; rsi now points to the next location
dec rdx ; Decrement the loop counter
jnz up1 ; Repeat for all 5 array elements
ret
你的'asci'函数是如何工作的？为什么不只是加'al'，'0'，而不是你的旋转/掩码/比较/'加'al'，'7'？另外，在普通的ABI中，rbx是保存的，但是你可以使用bl作为临时寄存器。你确定你没有写出数组的最后？例如**也许你第二次调用'asci'会从第一次调用结束，因为你只使用了10个字节的输出缓冲区。**我还没有弄清楚你的代码在哪里检测数组的长度（这取决于重叠的量）。 –
我已经尝试过将'al'，'0''转换为ASCII码，但问题在于我使用的是数组。此外，如果我使用该代码，它会显示分段错误。所以我用上面的书面程序转换成ASCII。我想你要明白，ASCII功能是如何工作的，如果是这样的话然后再看看节目，我已经添加了注释，使其更好地理解。现在针对你的第二个问题，你是否建议在我的第一个ASCII调用之后使用系统调用？ – Paras
哦，对了，转换为十六进制数字需要在'0-9'与'A-F'一个分支，我明白这是关于现在你已经改善了意见。对于缓冲区处理：如果您转换/打印，转换/打印，而不是转换，转换/打印，打印，则不需要两个单独的ASCII缓冲区。但是你仍然需要保留适量的空间，就像Score_Under的答案指出的那样。 –