使用C语言1.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

以下是使用C语言实现的计算器，通过栈实现中缀表达式求值：

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h> // 包含 isdigit() 函数

#define MAX_STACK_SIZE 100 // 栈的最大容量

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

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

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

int is_full(Stack *s) {
    return s->top == MAX_STACK_SIZE - 1;
}

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

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

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

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

int priority(char op) {
    if (op == '+' || op == '-')
        return 1;
    if (op == '*' || op == '/')
        return 2;
    return 0;
}

int evaluate(char *expr) {
    Stack operand_stack;
    Stack operator_stack;
    init(&operand_stack);
    init(&operator_stack);
    int i = 0;
    while (expr[i] != '\0') {
        if (isdigit(expr[i])) {
            int value = 0;
            while (isdigit(expr[i])) {
                value = value * 10 + expr[i] - '0';
                i++;
            }
            push(&operand_stack, value);
        } else if (is_operator(expr[i])) {
            while (!is_empty(&operator_stack) && priority(peek(&operator_stack)) >= priority(expr[i])) {
                int op = pop(&operator_stack);
                int op1 = pop(&operand_stack);
                int op2 = pop(&operand_stack);
                int result;
                switch (op) {
                    case '+':
                        result = op2 + op1;
                        break;
                    case '-':
                        result = op2 - op1;
                        break;
                    case '*':
                        result = op2 * op1;
                        break;
                    case '/':
                        result = op2 / op1;
                        break;
                }
                push(&operand_stack, result);
            }
            push(&operator_stack, expr[i]);
            i++;
        } else if (expr[i] == '(') {
            push(&operator_stack, expr[i]);
            i++;
        } else if (expr[i] == ')') {
            while (peek(&operator_stack) != '(') {
                int op = pop(&operator_stack);
                int op1 = pop(&operand_stack);
                int op2 = pop(&operand_stack);
                int result;
                switch (op) {
                    case '+':
                        result = op2 + op1;
                        break;
                    case '-':
                        result = op2 - op1;
                        break;
                    case '*':
                        result = op2 * op1;
                        break;
                    case '/':
                        result = op2 / op1;
                        break;
                }
                push(&operand_stack, result);
            }
            pop(&operator_stack);
            i++;
        } else {
            printf("Invalid input.\n");
            exit(1);
        }
    }
    while (!is_empty(&operator_stack)) {
        int op = pop(&operator_stack);
        int op1 = pop(&operand_stack);
        int op2 = pop(&operand_stack);
        int result;
        switch (op) {
            case '+':
                result = op2 + op1;
                break;
            case '-':
                result = op2 - op1;
                break;
            case '*':
                result = op2 * op1;
                break;
            case '/':
                result = op2 / op1;
                break;
        }
        push(&operand_stack, result);
    }
    return peek(&operand_stack);
}

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

使用上述代码，你可以输入一个中缀表达式，然后程序会计算并输出结果。例如，输入 `5*(10-3)/7`，输出 `Result: 2`。