1桁同士の四則演算 ref: http://qiita.com/items/916302df2fb2fe2e6a5c

math.c

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double bekijo(char str[]) {
    int operator, count = 0;
    double num1, num2;
    for (int i = 0; str[i] != '\0'; i++) {
        if (str[i] == '+') {
            operator = 0;
            count++;
        } else if (str[i] == '-') {
            operator = 1;
            count++;
        } else if (str[i] == '*') {
            operator = 2;
            count++;
        } else if (str[i] == '/') {
            operator = 3;
            count++;
        } else {
            if (count != 1) {
                num1 = atof(&str[i]);
            } else {
                num2 = atof(&str[i]);
            }
        }
    }
    if (operator == 0) {
        return num1 + num2;
    } else if (operator == 1) {
        return num1 - num2;
    } else if (operator == 2) {
        return num1 * num2;
    } else {
        return num1 / num2;
    }
}

int main(int argc, char *argv[]) {
    char str[256] = "5/2";
    printf("%s = %lf\n", str, bekijo(str));
    return 0;
}