输入一个中缀算术表达式，求解表达式的值。运算符包括+、-、*、/、(、)、=，使用c++编写代码

我可以为您提供一个简单的中缀表达式求解程序，以下是使用C语言编写的代码：

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

#define STACK_SIZE 50

// 符号栈结构体
struct stack {
    int items[STACK_SIZE];
    int top;
};

// 初始化符号栈
void init_stack(struct stack *s) {
    s->top = -1;
}

// 判断符号栈是否为空
int is_empty(struct stack *s) {
    return s->top == -1;
}

// 判断符号栈是否已满
int is_full(struct stack *s) {
    return s->top == STACK_SIZE - 1;
}

// 入栈操作
void push(struct stack *s, int item) {
    if (is_full(s)) {
        printf("\nStack is full!");
        exit(1);
    }
    s->items[++s->top] = item;
}

// 出栈操作
int pop(struct stack *s) {
    if (is_empty(s)) {
        printf("\nInvalid expression!");
        exit(1);
    }
    return s->items[s->top--];
}

// 获取栈顶元素
int peek(struct stack *s) {
    if (is_empty(s)) {
        printf("\nInvalid expression!");
        exit(1);
    }
    return s->items[s->top];
}

// 判断是否为操作符
int is_operator(char c) {
    return (c == '+' || c == '-' || c == '*' || c == '/' || c == '(' || c == ')');
}

// 判断操作符的优先级
int precedence(char c) {
    if (c == '(' || c == ')') {
        return 0;
    }
    if (c == '+' || c == '-') {
        return 1;
    }
    if (c == '*' || c == '/') {
        return 2;
    }
    return 0;
}

// 运算操作
int operate(int op1, int op2, char operator) {
    switch (operator) {
        case '+':
            return op1 + op2;
        case '-':
            return op1 - op2;
        case '*':
            return op1 * op2;
        case '/':
            return op1 / op2;
    }
    return 0;
}

// 解析中缀表达式
int evaluate(char *expression) {
    struct stack operand_stack, operator_stack;
    int i, n, operand1, operand2;
    char token, operator;
    init_stack(&operand_stack);
    init_stack(&operator_stack);

    n = strlen(expression);
    for (i = 0; i < n; i++) {
        token = expression[i];
        // 如果是数字，压入操作数栈
        if (isdigit(token)) {
            push(&operand_stack, token - '0');
        }
        // 如果是左括号，压入操作符栈
        else if (token == '(') {
            push(&operator_stack, token);
        }
        // 如果是右括号，将操作符栈中左括号之间的操作符依次弹出，并进行运算，最后将左括号弹出
        else if (token == ')') {
            while (peek(&operator_stack) != '(') {
                operand2 = pop(&operand_stack);
                operand1 = pop(&operand_stack);
                operator = pop(&operator_stack);
                push(&operand_stack, operate(operand1, operand2, operator));
            }
            pop(&operator_stack);
        }
        // 如果是操作符，需要进行进一步判断
        else if (is_operator(token)) {
            // 如果当前操作符优先级小于等于操作符栈顶的操作符，将操作符栈顶的操作符依次弹出，并进行运算
            while (!is_empty(&operator_stack) && precedence(token) <= precedence(peek(&operator_stack))) {
                operand2 = pop(&operand_stack);
                operand1 = pop(&operand_stack);
                operator = pop(&operator_stack);
                push(&operand_stack, operate(operand1, operand2, operator));
            }
            // 将当前操作符压入操作符栈
            push(&operator_stack, token);
        }
    }

    // 处理剩余的操作符
    while (!is_empty(&operator_stack)) {
        operand2 = pop(&operand_stack);
        operand1 = pop(&operand_stack);
        operator = pop(&operator_stack);
        push(&operand_stack, operate(operand1, operand2, operator));
    }

    // 返回最终结果
    return pop(&operand_stack);
}

int main() {
    char expression[50];
    int result;

    printf("Please input an infix arithmetic expression:\n");
    scanf("%s", expression);

    result = evaluate(expression);

    printf("\nResult = %d", result);

    return 0;
}

使用该程序求解中缀表达式的值时，需要注意输入表达式时需用括号括起来，例如：(1+2)*3。