拆分字符串程序集AT＆T

Question

如何拆分字符串,程序集中的分隔符AT & T？

這是我有字符串：6,3,1，2431，7，9，1221

欲1添加到所有的,之間的數字。

的結果應該是：7, 4, 2432, 8, 10, 1222

我與ubuntu14工作 - 64位，Intel的CPU和GAS編譯

Answer 1

僞代碼：

; si = input string 
di = output string 
if 0 == [si] jmp NoNumberFound 
NextNumberLoop: 
rax = 0 
ReadLoop: 
bl = [si] & inc si 
if 0 == bl jmp LastNumberFound 
if ',' == bl jmp CommaFound 
rax = rax*10 + (bl-'0') ; extend bl to 64b of course 
jmp ReadLoop 

CommaFound: 
call StoreNumberPlusOne 
[di]=',' & inc di 
jmp NextNumberLoop 

LastNumberFound: 
call StoreNumberPlusOne 
NoNumberFound: 
[di]='0' 
output resulting string and exit. 

StoreNumberPlusOne: 
inc rax (number+1) 
print decimal formatted rax to [di] + make di point after it 
ret 

（DI/SI 64b上平臺指針是當然rsi/rdi等等......它只是顯示算法的僞代碼，不能從字面上理解）

---

其他選項是在字符串本身進行分析而不用解析數字。

分配足夠的緩存生成的字符串（如果你把輸入作爲n次9，輸出緩衝器將幾乎兩倍只要輸入緩衝器，包含n次10）。

將輸入字符串複製到outputBuffer，並在其最後一個字符處具有指針ptr。

doIncrement = true 
while (outputBuffer <= ptr) { 
    if ',' == [ptr] { 
    if doIncrement { 
     move by 1 byte further everything from ptr+1 onward: 
     in ASM you can do that by simple REP MOVSB, but as the 
     source overlap destination, make sure you use STD (DF=1) 
     together with end pointers. 
     [ptr+1] = '1' 
    } 
    doIncrement = true 
    } else { 
    ; [ptr] should be between '0' to '9' (for valid input) 
    if doIncrement { 
     ++[ptr] 
     if '9'+1 == [ptr] { 
     [ptr] = '0' 
     } else { 
     doIncrement = false 
     } 
    } 
    } 
    --ptr; 
}