c语言栈实现四则运算

实现一个栈来进行四则运算，首先需要定义栈的数据结构。栈是一种后进先出的数据结构，可以使用数组或链表来实现。这里我们使用数组来实现。

#define MAX_STACK_SIZE 100

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 x)
{
    if (is_full(s)) {
        printf("Stack overflow\n");
        exit(1);
    }
    s->data[++s->top] = x;
}

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

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

接下来，我们可以使用这个栈来实现四则运算。我们假设输入的表达式中只包含加减乘除和数字，没有括号和负数。

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

#define MAX_EXPR_SIZE 100

int evaluate(char *expr)
{
    Stack operand;
    Stack operator;
    init(&operand);
    init(&operator);

    while (*expr != '\0') {
        if (isdigit(*expr)) {
            int num = 0;
            while (isdigit(*expr)) {
                num = num * 10 + (*expr - '0');
                expr++;
            }
            push(&operand, num);
        } else if (*expr == '+' || *expr == '-' || *expr == '*' || *expr == '/') {
            while (!is_empty(&operator) && peek(&operator) != '(' && ((*expr == '+' || *expr == '-') ? peek(&operator) != '*' && peek(&operator) != '/' : 1)) {
                int op = pop(&operator);
                int b = pop(&operand);
                int a = pop(&operand);
                switch (op) {
                    case '+':
                        push(&operand, a + b);
                        break;
                    case '-':
                        push(&operand, a - b);
                        break;
                    case '*':
                        push(&operand, a * b);
                        break;
                    case '/':
                        push(&operand, a / b);
                        break;
                }
            }
            push(&operator, *expr);
            expr++;
        } else if (*expr == '(') {
            push(&operator, *expr);
            expr++;
        } else if (*expr == ')') {
            while (peek(&operator) != '(') {
                int op = pop(&operator);
                int b = pop(&operand);
                int a = pop(&operand);
                switch (op) {
                    case '+':
                        push(&operand, a + b);
                        break;
                    case '-':
                        push(&operand, a - b);
                        break;
                    case '*':
                        push(&operand, a * b);
                        break;
                    case '/':
                        push(&operand, a / b);
                        break;
                }
            }
            pop(&operator);
            expr++;
        } else {
            printf("Invalid character: %c\n", *expr);
            exit(1);
        }
    }

    while (!is_empty(&operator)) {
        int op = pop(&operator);
        int b = pop(&operand);
        int a = pop(&operand);
        switch (op) {
            case '+':
                push(&operand, a + b);
                break;
            case '-':
                push(&operand, a - b);
                break;
            case '*':
                push(&operand, a * b);
                break;
            case '/':
                push(&operand, a / b);
                break;
        }
    }

    int result = pop(&operand);
    if (!is_empty(&operand)) {
        printf("Invalid expression\n");
        exit(1);
    }

    return result;
}

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

这个程序可以处理如下的表达式：

3+4*5-6/2
(3+4)*5/(2+3)