中缀表达式求值c++代码

我可以回答这个问题。中缀表达式求值可以通过将中缀表达式转换为后缀表达式，然后再用栈来计算后缀表达式的值。以下是一个示例 C 代码：

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

#define MAX_SIZE 100

typedef struct {
    int top;
    int data[MAX_SIZE];
} Stack;

void push(Stack *s, int x) {
    if (s->top == MAX_SIZE - 1) {
        printf("Stack overflow\n");
        exit(1);
    }
    s->data[++s->top] = x;
}

int pop(Stack *s) {
    if (s->top == -1) {
        printf("Stack underflow\n");
        exit(1);
    }
    return s->data[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 ;
    }
}

int evaluate(char *expression) {
    Stack s;
    s.top = -1;
    int i, j, len, a, b, result;
    char c, postfix[MAX_SIZE];
    len = strlen(expression);
    j = ;
    for (i = ; i < len; i++) {
        c = expression[i];
        if (c >= '' && c <= '9') {
            postfix[j++] = c;
        } else if (is_operator(c)) {
            while (s.top != -1 && is_operator(s.data[s.top]) && precedence(s.data[s.top]) >= precedence(c)) {
                postfix[j++] = pop(&s);
            }
            push(&s, c);
        } else if (c == '(') {
            push(&s, c);
        } else if (c == ')') {
            while (s.top != -1 && s.data[s.top] != '(') {
                postfix[j++] = pop(&s);
            }
            if (s.top == -1) {
                printf("Unmatched parentheses\n");
                exit(1);
            }
            pop(&s);
        } else {
            printf("Invalid character: %c\n", c);
            exit(1);
        }
    }
    while (s.top != -1) {
        if (s.data[s.top] == '(') {
            printf("Unmatched parentheses\n");
            exit(1);
        }
        postfix[j++] = pop(&s);
    }
    postfix[j] = '\';
    len = strlen(postfix);
    s.top = -1;
    for (i = ; i < len; i++) {
        c = postfix[i];
        if (c >= '' && c <= '9') {
            push(&s, c - '');
        } else if (is_operator(c)) {
            b = pop(&s);
            a = pop(&s);
            switch (c) {
                case '+':
                    result = a + b;
                    break;
                case '-':
                    result = a - b;
                    break;
                case '*':
                    result = a * b;
                    break;
                case '/':
                    result = a / b;
                    break;
            }
            push(&s, result);
        } else {
            printf("Invalid character: %c\n", c);
            exit(1);
        }
    }
    if (s.top != ) {
        printf("Invalid expression\n");
        exit(1);
    }
    return pop(&s);
}

int main() {
    char expression[MAX_SIZE];
    printf("Enter an infix expression: ");
    scanf("%s", expression);
    int result = evaluate(expression);
    printf("Result: %d\n", result);
    return ;
}