请用c++实现一个括号栈并判断括号是否匹配

好的，以下是用 C 实现一个括号栈并判断括号是否匹配的代码：

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

#define MAX_SIZE 100

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

void init(Stack *s) {
    s->top = -1;
}

bool is_empty(Stack *s) {
    return s->top == -1;
}

bool is_full(Stack *s) {
    return s->top == MAX_SIZE - 1;
}

void push(Stack *s, char c) {
    if (is_full(s)) {
        printf("Stack overflow!\n");
        exit(EXIT_FAILURE);
    }
    s->data[++s->top] = c;
}

char pop(Stack *s) {
    if (is_empty(s)) {
        printf("Stack underflow!\n");
        exit(EXIT_FAILURE);
    }
    return s->data[s->top--];
}

bool is_match(char left, char right) {
    return (left == '(' && right == ')') || (left == '[' && right == ']') || (left == '{' && right == '}');
}

bool is_balanced(char *expr) {
    Stack s;
    init(&s);
    for (int i = 0; expr[i] != '\0'; i++) {
        if (expr[i] == '(' || expr[i] == '[' || expr[i] == '{') {
            push(&s, expr[i]);
        } else if (expr[i] == ')' || expr[i] == ']' || expr[i] == '}') {
            if (is_empty(&s) || !is_match(pop(&s), expr[i])) {
                return false;
            }
        }
    }
    return is_empty(&s);
}

int main() {
    char expr[MAX_SIZE];
    printf("Enter an expression: ");
    scanf("%s", expr);
    if (is_balanced(expr)) {
        printf("The expression is balanced.\n");
    } else {
        printf("The expression is not balanced.\n");
    }
    return 0;
}

如果输入的表达式是平衡的，程序会输出"The expression is balanced."，否则会输出"The expression is not balanced."。

希望这个回答能够帮到你！