1桁同士の四則演算

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;
}