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Super simple calculator
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| #include <stdio.h> | |
| #include <ctype.h> | |
| #include <stdlib.h> | |
| #include <limits.h> | |
| enum RET { | |
| RET_OK = 0, | |
| RET_OVERFLOW = -1, | |
| RET_NODIGIT = -2, | |
| }; | |
| static char* _skip_space(char *str) | |
| { | |
| while (isspace(*str)) | |
| str++; | |
| return str; | |
| } | |
| static enum RET _get_num(const char *str, | |
| char **endptr, | |
| long int *res) | |
| { | |
| long int cur_val; | |
| cur_val = strtol(str, endptr, 0); | |
| /* Check overflow. */ | |
| if (cur_val == LONG_MAX || cur_val == LONG_MIN) | |
| { | |
| printf("overflow at: %s\n", str); | |
| return RET_OVERFLOW; | |
| } | |
| /* No digits found. */ | |
| if (*endptr == str) | |
| { | |
| printf("no digit at: %s\n", str); | |
| return RET_NODIGIT; | |
| } | |
| *res = cur_val; | |
| return RET_OK; | |
| } | |
| /* return the calculate status. 0 is OK. | |
| * | |
| * A block is a formula surrounded by (). | |
| */ | |
| static enum RET _calc_block(char *str, | |
| char **endptr, | |
| long int *res) | |
| { | |
| enum RET ret; | |
| long int cur_val; | |
| long int next_val; | |
| str = _skip_space(str); | |
| if (*str == '(') | |
| { | |
| /* We get here because the last node is +- or the first char is '('. So | |
| * we need to proceed parsing as there may be other operations behind. | |
| * In the a*(b+c)-d condition, the following operation is handled by | |
| * the handler of *. */ | |
| ret = _calc_block(str + 1, endptr, &cur_val); | |
| if (ret != RET_OK) | |
| return ret; | |
| goto _parse_next; | |
| } | |
| ret = _get_num(str, endptr, &cur_val); | |
| if (ret != RET_OK) | |
| { | |
| return ret; | |
| } | |
| _parse_next: | |
| str = _skip_space(*endptr); | |
| switch (*str) { | |
| case '+': { | |
| /* a + b [+/-*] c is equal to a + (b [+/-*] c) */ | |
| ret = _calc_block(str+1, endptr, &next_val); | |
| if (ret != 0) | |
| { | |
| return ret; | |
| } | |
| *res = cur_val + next_val; | |
| return RET_OK; | |
| } | |
| break; | |
| case '-': { | |
| /* a - b [+/-*] c is equal to a + (b [-/+*] c) */ | |
| char *p; | |
| for (p = str + 1; *p != '\0'; p++) | |
| { | |
| if (*p == '-') | |
| *p = '+'; | |
| else if (*p == '+') | |
| *p = '-'; | |
| } | |
| ret = _calc_block(str+1, endptr, &next_val); | |
| if (ret != 0) | |
| { | |
| return ret; | |
| } | |
| *res = cur_val - next_val; | |
| return RET_OK; | |
| } | |
| break; | |
| case '*': case '/': { | |
| /* a * b [+-] c is not equal to a * (b [+-] c). So we have to handle it in line. */ | |
| /* Store current operator as we need to modify @str. */ | |
| char op = *str; | |
| str = _skip_space(str + 1); | |
| if (*str == '(') | |
| ret = _calc_block(str + 1, endptr, &next_val); | |
| else | |
| ret = _get_num(str, endptr, &next_val); | |
| if (ret != 0) | |
| { | |
| return ret; | |
| } | |
| if (op == '*') | |
| cur_val *= next_val; | |
| else | |
| cur_val /= next_val; | |
| /* Continue parsing. */ | |
| goto _parse_next; | |
| } | |
| break; | |
| case '\0': case ')': | |
| *res = cur_val; | |
| *endptr = (char*)(str + 1); | |
| return RET_OK; | |
| default: | |
| printf("unkown token at: %s\n", str); | |
| break; | |
| }; | |
| return -1; | |
| } | |
| int calc(char *str, long int *res) | |
| { | |
| char *ptr; | |
| return _calc_block(str, &ptr, res); | |
| } | |
| #include <string.h> | |
| int main(int argc, char * argv[]) | |
| { | |
| int ret; | |
| long int res = 0; | |
| char *ostr = strdup(argv[1]); | |
| ret = calc(argv[1], &res); | |
| printf("calc %s: %ld\n", ostr, res); | |
| return ret; | |
| } |
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| #!/bin/sh | |
| set -e | |
| gcc -Wall calc.c | |
| test() { | |
| res=$(echo "$1" | bc) | |
| echo "calc $1: $res" | |
| ./a.out "$1" | |
| } | |
| test '1+1' | |
| test '1-1' | |
| test '1*1' | |
| test '1+(1+4)' | |
| test '1-1-4' | |
| test '1-1*4' | |
| test '1-1*4-9' | |
| test '1-1*4-9+7' | |
| test '(1+1)*7' | |
| test '1+1*7' | |
| test '8/1/2/2' | |
| test '8*(1+3)+9' | |
| test '1*(1+2)+9' | |
| test '1*(1*2)*9' | |
| test '8*(1+3)*9' | |
| test '8+(1+3)*9' | |
| test '8*(1+3)+9' | |
| test '1+(1+2)+9' | |
| test '1+1+(2+9)' | |
| test '3*(1+2+9)' | |
| test '(3+1)*2+9' |
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暂时不支持 ()。暂时只支持整形(strtof 的错误处理稍微麻烦些，就没有做)。