数据结构栈的算术表达式求值C语言

栈是一种后进先出（LIFO）的数据结构，它可以用来解决算术表达式求值问题。下面是一个使用栈求解算术表达式的C语言程序：

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

#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];
}

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

    while (*expr != '\0') {
        if (isdigit(*expr)) {
            int value = 0;
            while (isdigit(*expr)) {
                value = value * 10 + (*expr - '0');
                expr++;
            }
            push(&operand_stack, value);
        } else if (*expr == '(') {
            push(&operator_stack, *expr);
            expr++;
        } else if (*expr == ')') {
            while (peek(&operator_stack) != '(') {
                int op2 = pop(&operand_stack);
                int op1 = pop(&operand_stack);
                char op = pop(&operator_stack);
                int result;
                switch (op) {
                    case '+': result = op1 + op2; break;
                    case '-': result = op1 - op2; break;
                    case '*': result = op1 * op2; break;
                    case '/': result = op1 / op2; break;
                }
                push(&operand_stack, result);
            }
            pop(&operator_stack);
            expr++;
        } else if (*expr == '+' || *expr == '-' || *expr == '*' || *expr == '/') {
            while (!is_empty(&operator_stack) && peek(&operator_stack) != '(' && ((*expr == '+' || *expr == '-') && (peek(&operator_stack) == '*' || peek(&operator_stack) == '/'))) {
                int op2 = pop(&operand_stack);
                int op1 = pop(&operand_stack);
                char op = pop(&operator_stack);
                int result;
                switch (op) {
                    case '+': result = op1 + op2; break;
                    case '-': result = op1 - op2; break;
                    case '*': result = op1 * op2; break;
                    case '/': result = op1 / op2; break;
                }
                push(&operand_stack, result);
            }
            push(&operator_stack, *expr);
            expr++;
        } else {
            printf("Invalid character: %c\n", *expr);
            exit(1);
        }
    }

    while (!is_empty(&operator_stack)) {
        int op2 = pop(&operand_stack);
        int op1 = pop(&operand_stack);
        char op = pop(&operator_stack);
        int result;
        switch (op) {
            case '+': result = op1 + op2; break;
            case '-': result = op1 - op2; break;
            case '*': result = op1 * op2; break;
            case '/': result = op1 / op2; break;
        }
        push(&operand_stack, result);
    }

    return pop(&operand_stack);
}

int main() {
    char expr[100];
    printf("Enter an arithmetic expression: ");
    scanf("%s", expr);
    int result = evaluate(expr);
    printf("Result: %d\n", result);
    return 0;
}

该程序使用两个栈，一个用于存储操作数，另一个用于存储运算符。它遍历算术表达式中的每个字符，并根据字符的类型执行相应的操作。如果字符是数字，则将其转换为整数并将其推入操作数栈中；如果字符是左括号，则将其推入运算符栈中；如果字符是右括号，则从运算符栈中弹出运算符和操作数，执行相应的运算，并将结果推入操作数栈中，直到遇到左括号；如果字符是加、减、乘或除号，则从运算符栈中弹出运算符和操作数，执行相应的运算，并将结果推入操作数栈中，直到运算符栈为空或栈顶运算符的优先级低于当前运算符的优先级。最后，操作数栈中剩下的唯一元素就是算术表达式的值。