c实现中缀表达式转后缀表达式代码

以下是C语言实现中缀表达式转后缀表达式的代码，同时也包含了计算后缀表达式的功能：

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

#define MAX_STACK_SIZE 100

typedef struct {
    int top;
    char items[MAX_STACK_SIZE];
} Stack;

void push(Stack *s, char c) {
    if (s->top == MAX_STACK_SIZE - 1) {
        printf("Stack is full\n");
        exit(1);
    }
    s->items[++(s->top)] = c;
}

char pop(Stack *s) {
    if (s->top == -1) {
        printf("Stack is empty\n");
        exit(1);
    }
    return s->items[(s->top)--];
}

int is_operator(char c) {
    return (c == '+' || c == '-' || c == '*' || c == '/');
}

int precedence(char c) {
    if (c == '+' || c == '-') {
        return 1;
    } else if (c == '*' || c == '/') {
        return 2;
    } else {
        return 0;
    }
}

void infix_to_postfix(char infix[], char postfix[]) {
    Stack s;
    s.top = -1;
    int i = 0, j = 0;
    char c, temp;
    while ((c = infix[i++]) != '\0') {
        if (isdigit(c)) {
            postfix[j++] = c;
        } else if (is_operator(c)) {
            while (s.top != -1 && precedence(s.items[s.top]) >= precedence(c)) {
                postfix[j++] = pop(&s);
            }
            push(&s, c);
        } else if (c == '(') {
            push(&s, c);
        } else if (c == ')') {
            while ((temp = pop(&s)) != '(') {
                postfix[j++] = temp;
            }
        }
    }
    while (s.top != -1) {
        postfix[j++] = pop(&s);
    }
    postfix[j] = '\0';
}

int evaluate_postfix(char postfix[]) {
    Stack s;
    s.top = -1;
    int i = 0, op1, op2, result;
    char c;
    while ((c = postfix[i++]) != '\0') {
        if (isdigit(c)) {
            push(&s, c - '0');
        } else if (is_operator(c)) {
            op2 = pop(&s);
            op1 = pop(&s);
            switch (c) {
                case '+':
                    result = op1 + op2;
                    break;
                case '-':
                    result = op1 - op2;
                    break;
                case '*':
                    result = op1 * op2;
                    break;
                case '/':
                    result = op1 / op2;
                    break;
            }
            push(&s, result);
        }
    }
    return pop(&s);
}

int main() {
    char infix[100], postfix[100];
    printf("Enter infix expression: ");
    scanf("%s", infix);
    infix_to_postfix(infix, postfix);
    printf("Postfix expression: %s\n", postfix);
    printf("Result: %d\n", evaluate_postfix(postfix));
    return 0;
}