C++经典算法题-后序式的运算

1	1
2	1 2
+	3 // 1+2 后存回
3	3 3
4	3 3 4
+	3 7 // 3+4 后存回
*	21 // 3 * 7  后存回

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

    void evalPf(char*);

    double cal(double, char, double);

    int main(void) {
        char input[ 80];

        printf("输入后序式：");
        scanf("%s", input);
        evalPf(input);
        return 0;
    }

    void evalPf(char*postfix) {
        double stack[ 80] ={
            0.0
        } ;
        char temp[ 2];
        char token;
        int top = 0, i = 0;
        temp[1] = '\0';
        while (1) {
            token = postfix[i];
            switch (token) {
                case '\0':
                    printf("ans = %f\n", stack[top]);
                    return;
                case '+':
                case '-':
                case '*':
                case '/':
                    stack[top - 1] =
                            cal(stack[top], token, stack[top - 1]);
                    top--;
                    break;
                default:
                    if (top < sizeof(stack) / sizeof( float)){
                    temp[0] = postfix[i];
                    top++;
                    stack[top] = atof(temp);


                }
                i++;
            }
        }

    }
	break;


    double cal(double p1, char op, double p2) {
        switch (op) {
            case '+':
                return p1 + p2;

            case '-':
                return p1 - p2;
            case '*':
                return p1 * p2;
            case '/':
                return p1 / p2;
        }
    }