用C语言写一个可以处理括号内运算的四则运算计算器

好的，以下是一个可以处理括号内运算的四则运算计算器的C语言代码：

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

#define STACK_INIT_SIZE 20
#define STACKINCREMENT 10
#define MAXBUFFER 10

typedef struct
{
    double *base;
    double *top;
    int stacksize;
} Stack;

void InitStack(Stack *S)
{
    S->base = (double *)malloc(STACK_INIT_SIZE * sizeof(double));
    if (!S->base)
        exit(0);
    S->top = S->base;
    S->stacksize = STACK_INIT_SIZE;
}

void Push(Stack *S, double e)
{
    if (S->top - S->base >= S->stacksize)
    {
        S->base = (double *)realloc(S->base, (S->stacksize + STACKINCREMENT) * sizeof(double));
        if (!S->base)
            exit(0);
        S->top = S->base + S->stacksize;
        S->stacksize += STACKINCREMENT;
    }
    *(S->top) = e;
    S->top++;
}

void Pop(Stack *S, double *e)
{
    if (S->top == S->base)
        return;
    *e = *(--S->top);
}

double GetTop(Stack *S)
{
    if (S->top != S->base)
        return *(S->top - 1);
    return 0;
}

int StackEmpty(Stack *S)
{
    if (S->top == S->base)
        return 1;
    else
        return 0;
}

int OperatorPriority(char op)
{
    switch (op)
    {
    case '+':
    case '-':
        return 1;
    case '*':
    case '/':
        return 2;
    case '(':
        return 0;
    }
    return -1;
}

double Calculate(double a, char op, double b)
{
    switch (op)
    {
    case '+':
        return a + b;
    case '-':
        return a - b;
    case '*':
        return a * b;
    case '/':
        return a / b;
    }
    return 0;
}

double EvaluateExpression(char *exp)
{
    Stack numStack;
    Stack opStack;
    double num = 0;
    char buffer[MAXBUFFER];
    int i = 0;
    InitStack(&numStack);
    InitStack(&opStack);
    while (*exp != '\0')
    {
        if (isdigit(*exp) || *exp == '.')
        {
            buffer[i++] = *exp;
            if (i >= MAXBUFFER)
                exit(0);
        }
        else if (*exp == '+' || *exp == '-' || *exp == '*' || *exp == '/')
        {
            buffer[i] = '\0';
            num = atof(buffer);
            i = 0;
            while (!StackEmpty(&opStack) && OperatorPriority(*exp) <= OperatorPriority(GetTop(&opStack)))
            {
                double b, a;
                Pop(&numStack, &b);
                Pop(&numStack, &a);
                Pop(&opStack, &opStack);
                Push(&numStack, Calculate(a, opStack, b));
            }
            Push(&opStack, *exp);
        }
        else if (*exp == '(')
        {
            Push(&opStack, *exp);
        }
        else if (*exp == ')')
        {
            buffer[i] = '\0';
            num = atof(buffer);
            i = 0;
            while (!StackEmpty(&opStack) && GetTop(&opStack) != '(')
            {
                double b, a;
                Pop(&numStack, &b);
                Pop(&numStack, &a);
                Pop(&opStack, &opStack);
                Push(&numStack, Calculate(a, opStack, b));
            }
            if (GetTop(&opStack) == '(')
            {
                Pop(&opStack, &opStack);
            }
        }
        exp++;
    }
    buffer[i] = '\0';
    num = atof(buffer);
    while (!StackEmpty(&opStack))
    {
        double b, a;
        Pop(&numStack, &b);
        Pop(&numStack, &a);
        Pop(&opStack, &opStack);
        Push(&numStack, Calculate(a, opStack, b));
    }
    if (!StackEmpty(&numStack))
    {
        double result;
        Pop(&numStack, &result);
        return result;
    }
    return 0;
}

int main()
{
    char exp[100];
    printf("请输入四则运算表达式:");
    gets(exp);
    printf("结果为:%f", EvaluateExpression(exp));
    return 0;
}

这个计算器可以处理带有括号的四则运算表达式，支持加、减、乘、除和小数点。当然，还有很多可以改进和优化的地方，比如错误处理、表达式合法性检查等，但这里只是一个简单的演示，供参考。