如何在emu8086中计算字符串中特定字符的出现
问题描述：
请. .请在这个问题上帮助我.
Please. .kindly help me on this problem. .
输出:
输入字符串:已经
A-2
B-0
C-0
D-1
E-1
答
编写一个过程，该过程遍历整个字符串以查找特定字符.对于每个比赛，增加一个计数器.返回显示结果时，字符DL出现DH次:
例如A2".
Write a procedure that loops over the entire string in search of a specific character. For each match increment a counter. Upon return display the result as character DL occurs DH times:
e.g. "A - 2".
mov dl, "A"
call CountChar
... print result ...
mov dl, "B"
call CountChar
... print result ...
CountChar:
mov dh, 0
mov cx, ... length of the input string ...
jcxz Ready
mov bx, ... address of the input string ...
Again:
cmp [bx], dl
jne Skip
inc dh
Skip:
inc bx
loop Again
Ready:
ret