字符串的高级应用-char a[100] = "1+2=;3-2=;2*5=;8/4=;" 得到char a[100] ="1+2=3;3-2=1;2*5=10;8/4=2;"

 1 #include<stdio.h>
 2 #include<string.h>
 3
 4 int main()
 5 {
 6     char a[100] = "1+2=;3-2=;2*5=;8/4=;" ;
 7     char b[100] = {0};
 8     char *s;
 9     s = strtok(a, ";");
10     while(s)
11     {
12
13         int i,j;
14         char c;
15         sscanf(s, "%d%c%d=", &i, &c, &j);
16         int res = 0;
17         switch(c)
18         {
19         case ‘+‘:
20             res = i + j;
21             break;
22
23         case ‘-‘:
24             res = i - j;
25             break;
26
27         case ‘*‘:
28             res = i * j;
29             break;
30
31         case ‘/‘:
32             res = i / j;
33             break;
34         default    :
35             res = 0;
36
37         }
38
39         char tmp[100] = {0};
40         sprintf(tmp, "%s%d;", s, res);   //把运算式放入到tmp数组
41         strcat(b, tmp);    //tmp是在循环内定义的，因此要在循环外再定义一个b
42         s = strtok(NULL, ";");
43
44     }
45
46         strcpy(a, b);
47         printf("%s",a);
48     return 0;
49
50  } 

原文地址：https://www.cnblogs.com/yangxiaoqin/p/8371754.html