c语言求带括号的四则运算表达式，给出一份完整代码

在C语言中，处理带括号的四则运算表达式可以使用递归解析法，利用栈的数据结构来保存操作数和运算符。下面是一个简单的例子，使用了递归函数`calculate()`来解决这个问题。请注意，这个示例仅适用于包含加减乘除和小括号的简单算术表达式，对于更复杂的数学库或者需要支持浮点数、优先级判断等功能的情况，可能需要引入正则表达式或编译原理的知识。

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

#define MAX_TOKENS 100

typedef enum {
    NUM,
    OPAREN,
    CPAREN,
    PLUS,
    MINUS,
    TIMES,
    DIVIDE
} Token;

char tokens[MAX_TOKENS][5];
int token_count = 0;

void tokenize(char *expression) {
    int i = 0;
    while (expression[i]) {
        if (expression[i] >= '0' && expression[i] <= '9') {
            tokens[token_count++] = "NUM";
            do {
                tokens[token_count - 1] += expression[i++];
            } while (expression[i] >= '0' && expression[i] <= '9');
        }
        else if (expression[i] == '(') {
            tokens[token_count++] = "OPAREN";
        }
        else if (expression[i] == ')') {
            tokens[token_count++] = "CPAREN";
        }
        else if (expression[i] == '+' || expression[i] == '-' ||
                 expression[i] == '*' || expression[i] == '/') {
            tokens[token_count++] = (expression[i] == '+') ? "PLUS" :
                                   (expression[i] == '-') ? "MINUS" :
                                   (expression[i] == '*') ? "TIMES" :
                                                      "DIVIDE";
            i++;
        }
        else {
            i++;
        }
    }
}

double calculate(char *expression) {
    double result = atof(tokens[0]); // Start with the first number
    for (int i = 1; i < token_count; i++) {
        double op2;
        char operator_ = tokens[i];

        if (operator_[0] != 'N') { // It's an operator
            op2 = atof(tokens[i + 1]);
            switch (operator_) {
                case 'PLUS':
                    result += op2;
                    break;
                case 'MINUS':
                    result -= op2;
                    break;
                case 'TIMES':
                    result *= op2;
                    break;
                case 'DIVIDE':
                    if (op2 != 0)
                        result /= op2;
                    else
                        return -1; // Division by zero error
                    break;
                default:
                    printf("Invalid operator: %s\n", tokens[i]);
                    return -1;
            }
            i++; // Skip the second operand
        }
        else { // It's a number
            result = atof(tokens[i]);
        }
    }

    return result;
}

int main() {
    char expression[] = "((3+4)*5)-2";
    tokenize(expression);
    printf("Result: %.2f\n", calculate(expression));
    return 0;
}