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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)。