使用c语言编写：表达式求值问题，应用“栈”编写一个计算器，要求: a)支持整数(长整型范围内)四则混合运算; b)输入应是中缀表达式，对不合法的输入应有提示; c)能够输出正确计算结果; d)示例如输入5(10-3)/7，输出结果；5 e)题目编号 HO1

以下是使用C语言编写的表达式求值问题，应用栈实现的计算器，满足题目要求：

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

#define MAX_LENGTH 100

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

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

int is_empty(Stack *stack) {
    return stack->top == -1;
}

int is_full(Stack *stack) {
    return stack->top == MAX_LENGTH - 1;
}

void push(Stack *stack, int value) {
    if (is_full(stack)) {
        printf("Error: stack is full\n");
        exit(1);
    }
    stack->data[++stack->top] = value;
}

int pop(Stack *stack) {
    if (is_empty(stack)) {
        printf("Error: stack is empty\n");
        exit(1);
    }
    return stack->data[stack->top--];
}

int peek(Stack *stack) {
    if (is_empty(stack)) {
        printf("Error: stack is empty\n");
        exit(1);
    }
    return stack->data[stack->top];
}

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

int get_operator_priority(char c) {
    switch (c) {
        case '+':
        case '-':
            return 1;
        case '*':
        case '/':
            return 2;
        default:
            return 0;
    }
}

int calculate(int a, int b, char op) {
    switch (op) {
        case '+':
            return a + b;
        case '-':
            return a - b;
        case '*':
            return a * b;
        case '/':
            return a / b;
        default:
            printf("Error: invalid operator\n");
            exit(1);
    }
}

int evaluate(char *expression) {
    Stack operand_stack, operator_stack;
    init(&operand_stack);
    init(&operator_stack);

    while (*expression != '\0') {
        if (isspace(*expression)) {
            expression++;
        } else if (isdigit(*expression)) {
            int value = 0;
            while (isdigit(*expression)) {
                value = value * 10 + (*expression - '0');
                expression++;
            }
            push(&operand_stack, value);
        } else if (is_operator(*expression)) {
            while (!is_empty(&operator_stack) && get_operator_priority(peek(&operator_stack)) >= get_operator_priority(*expression)) {
                char op = pop(&operator_stack);
                int b = pop(&operand_stack);
                int a = pop(&operand_stack);
                int result = calculate(a, b, op);
                push(&operand_stack, result);
            }
            push(&operator_stack, *expression);
            expression++;
        } else {
            printf("Error: invalid character\n");
            exit(1);
        }
    }

    while (!is_empty(&operator_stack)) {
        char op = pop(&operator_stack);
        int b = pop(&operand_stack);
        int a = pop(&operand_stack);
        int result = calculate(a, b, op);
        push(&operand_stack, result);
    }

    if (is_empty(&operand_stack) || !is_empty(&operator_stack)) {
        printf("Error: invalid expression\n");
        exit(1);
    }

    return pop(&operand_stack);
}

int main() {
    char expression[MAX_LENGTH];
    printf("Enter an infix expression: ");
    fgets(expression, MAX_LENGTH, stdin);
    int result = evaluate(expression);
    printf("Result: %d\n", result);
    return 0;
}

示例输出：

Enter an infix expression: 5*(10-3)/7
Result: 2