运算字符串解析（c语言）
运算字符的解析，如：43+表示：4+3， 25* 表示：2*5， 25*1+表示（2*5）+1，435/+表示：4 +（3/5）
1 #include <stdio.h> 2 #include <stdbool.h> 3 #include <string.h> 4 5 bool isNumber(const char *c); 6 int getNumber(const char *c); 7 bool isOperator(const char *c); 8 float calculate(char *string, int len); 9 10 int main(int argc, const char * argv[]) 11 { 12 13 // insert code here... 14 printf("Begin>>\n"); 15 printf("%d",'0'); 16 printf("%d\n",'/'); 17 char string[100] = "435/+"; 18 size_t len = strlen(string); 19 float ret = calculate(string, (int)len); 20 printf("result = %f\n", ret); 21 return 0; 22 } 23 24 bool isNumber(const char *c) 25 { 26 if ( (int)(*c) >= (int)('0') && (int)(*c) <= (int)('9')) { 27 return true; 28 } 29 return false; 30 } 31 32 int getNumber(const char *c) 33 { 34 bool isN = isNumber(c); 35 if (isN) { 36 int n = (int)((*c) - (int)'0'); 37 return n; 38 }else { 39 return 0; 40 } 41 } 42 43 bool isOperator(const char *c) 44 { 45 if ((int)(*c) == (int)'+' || (int)(*c) == (int)'-' || (int)(*c) == (int)('*') || (int)(*c) == (int)('/') ) { 46 return true; 47 } else { 48 return false; 49 } 50 51 } 52 53 float calculate(char *string, int len) 54 { 55 if (len == 1) { 56 return (float)getNumber(string); 57 } 58 59 if (len == 2) { 60 printf("error!"); 61 } 62 63 if (!isOperator(string + len - 1)) { 64 return 0.0f; 65 } 66 67 float a = 0.0f; 68 float b = 0.0f; 69 if (isOperator(string + len - 1 -1)) { 70 a = (float)getNumber(string); 71 b = calculate(string + 1, len - 2); 72 } else { 73 a = calculate(string, len - 2); 74 b = (float)getNumber(string + len - 1 - 1); 75 } 76 77 char operator = *(string + len - 1); 78 if ((int)operator == (int)'+') { 79 return (a + b); 80 } else if ((int)operator == (int)'-') { 81 return (a - b); 82 } else if ((int)operator == (int)'*') { 83 return (a * b); 84 } else if ((int)operator == (int)'/') { 85 return (a / b); 86 } else { 87 return 0.0f; 88 } 89 90 }
