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？

这是我有字符串：6,3,1，2431，7，9，1221

欲1添加到所有的,之间的数字。

的结果应该是：7, 4, 2432, 8, 10, 1222

我与ubuntu14工作 - 64位，Intel的CPU和GAS编译

Answer 1

伪代码：

; 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上平台指针是当然rsi/rdi等等......它只是显示算法的伪代码，不能从字面上理解）

---

其他选项是在字符串本身进行分析而不用解析数字。

分配足够的缓存生成的字符串（如果你把输入作为n次9，输出缓冲器将几乎两倍只要输入缓冲器，包含n次10）。

将输入字符串复制到outputBuffer，并在其最后一个字符处具有指针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; 
}