使用C语言实现中缀表达式的计算

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

typedef struct {
    int* data;
    int top;
    int capacity;
} Stack;

void initialize(Stack* stack, int capacity) {
    stack->data = (int*)malloc(capacity * sizeof(int));
    stack->top = -1;
    stack->capacity = capacity;
}

void push(Stack* stack, int element) {
    if (stack->top == stack->capacity - 1) {
        printf("栈已满，无法入栈\n");
    } else {
        stack->data[++stack->top] = element;
    }
}

int pop(Stack* stack) {
    if (stack->top == -1) {
        printf("栈为空，无法出栈\n");
        return -1;
    } else {
        return stack->data[stack->top--];
    }
}

int peek(Stack* stack) {
    if (stack->top != -1) {
        return stack->data[stack->top];
    }
    return -1;
}

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

int isOperator(char ch) {
    if (ch == '+' || ch == '-' || ch == '*' || ch == '/')
        return 1;
    else
        return 0;
}

int calculate(int operand1, int operand2, char operator) {
    switch (operator) {
        case '+':
            return operand1 + operand2;
        case '-':
            return operand1 - operand2;
        case '*':
            return operand1 * operand2;
        case '/':
            return operand1 / operand2;
        default:
            return 0;
    }
}

int evaluate(char* expression) {
    Stack operandStack;
    Stack operatorStack;
    int i = 0;

    initialize(&operandStack, 100);
    initialize(&operatorStack, 100);

    while (expression[i] != '\0') {
        if (expression[i] >= '0' && expression[i] <= '9') {
            int operand = 0;
            while (expression[i] >= '0' && expression[i] <= '9') {
                operand = operand * 10 + (expression[i] - '0');
                i++;
            }
            push(&operandStack, operand);
        } else if (isOperator(expression[i])) {
            while (!isEmpty(&operatorStack) && peek(&operatorStack) != '(' &&
                   peek(&operatorStack) != ')' && peek(&operatorStack) != '^' &&
                   peek(&operatorStack) != '*' && peek(&operatorStack) != '/') {
                int operand2 = pop(&operandStack);
                int operand1 = pop(&operandStack);
                char operator = pop(&operatorStack);
                int result = calculate(operand1, operand2, operator);
                push(&operandStack, result);
            }
            push(&operatorStack, expression[i]);
            i++;
        } else if (expression[i] == '(') {
            push(&operatorStack, expression[i]);
            i++;
        } else if (expression[i] == ')') {
            while (!isEmpty(&operatorStack) && peek(&operatorStack) != '(') {
                int operand2 = pop(&operandStack);
                int operand1 = pop(&operandStack);
                char operator = pop(&operatorStack);
                int result = calculate(operand1, operand2, operator);
                push(&operandStack, result);
            }
            pop(&operatorStack); // 弹出左括号
            i++;
        } else {
            i++;
        }
    }

    while (!isEmpty(&operatorStack)) {
        int operand2 = pop(&operandStack);
        int operand1 = pop(&operandStack);
        char operator = pop(&operatorStack);
        int result = calculate(operand1, operand2, operator);
        push(&operandStack, result);
    }

    int finalResult = pop(&operandStack);

    return finalResult;
}

int main() {
    char expression[100];

    printf("请输入中缀表达式：");
    fgets(expression, sizeof(expression), stdin);

    int result = evaluate(expression);

    printf("计算结果为：%d\n", result);

    return 0;
}