输入一个含两个正整数(int 类型)和一个操作符(+,-,*,/,%)的字符串(如:243/3)
a) 编写函数 int FindOp (char str, int nLen); 找到操作符(+,-,,/,%)在给定字符 串 str 中的位置, 返回其下标. nLen 为字符个数;
b) 编写函数 int FindDigitals (char *str, int nLen, int nStart, int nEnd), 给定起始 位置 nStart 和终止位置 nEnd, 从给定字符串 str 中提取子字符串, 并将其从字符串转换为 int 类型的数据返回. 如果终点位置 nEnd 为-1, 则子字符串范围 从 nStart 到字符串末尾. nLen 为字符个数;
c) 编写主函数, 实现功能: 输入字符串, 调用上述两个函数, 将字符串形式的计算式解析、计算得到结果, 并输出结果
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int FindOp (char*str, int nLen){
int i =0,n;
for(i=0;i<nLen;i++){
n=(int)str[i];
if(n<(int)'0' || n>(int)'9')
return i;
}
return -1;
}
int FindDigitals(char *str, int nLen, int nStart, int nEnd){
if(nEnd==-1){
nEnd = nLen - 1;
}
char*tmp=(char*)malloc(sizeof(char)*nLen);
tmp[0]='\0';
strncpy(tmp,str+nStart,nEnd-nStart+1);
int result = atoi(tmp);
free(tmp);
return result;
}
int main(){
char*test = "123+32";
char opt;
int len = strlen(test);
int index = FindOp(test,len);
opt = test[index];
int a = FindDigitals(test, len, 0, index-1),
b = FindDigitals(test, len, index+1, -1);
int result=0;
switch(opt){
case '+':
result = a+b;break;
case '-':
result = a-b;break;
case '*':
result = a*b;break;
case '/':
result = a/b;break;
case '%':
result = a%b;break;
}
printf("%d",result);
}
改动:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int FindOp(char* str, int nLen) {
int i = 0, n;
for (i = 0; i < nLen; i++) {
n = (int)str[i];
if (n<(int)'0' || n>(int)'9')
return i;
}
return -1;
}
int FindDigitals(char* str, int nLen, int nStart, int nEnd) {
if (nEnd == -1) {
nEnd = nLen - 1;
}
char* tmp = (char*)malloc(sizeof(char) * nLen);
if (tmp == NULL) exit(-1);
tmp[0] = '\0';
strncpy_s(tmp, nLen, str + nStart, nEnd - nStart + 1); //复制部分字符串
int result = atoi(tmp);
free(tmp);
return result;
}
int main() {
char test[10] = "123+32";
char opt;
int len = strlen(test);
int index = FindOp(test, len);
opt = test[index];
int a = FindDigitals(test, len, 0, index-1),
b = FindDigitals(test, len, index + 1, -1);
int result = 0;
switch (opt) {
case '+':
result = a + b; break;
case '-':
result = a - b; break;
case '*':
result = a * b; break;
case '/':
result = a / b; break;
case '%':
result = a % b; break;
}
printf("%d", result);
}