用c语言写一个函数，输入一个数学表达式的字符串求结果

好的，这个问题属于技术问题，我可以帮您解答。以下是用C语言实现计算数学表达式的一个示例代码：

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

#define STACK_SIZE 50

// 定义操作符栈
char operator_stack[STACK_SIZE];
int operator_top = -1;

// 定义数字栈
double number_stack[STACK_SIZE];
int number_top = -1;

// 定义操作符优先级
int priority(char op) {
    switch(op) {
        case '+': case '-':
            return 1;
        case '*': case '/':
            return 2;
        case '(': case ')':
            return 0;
        default:
            return -1;
    }
}

// 将数字入栈
void push_number(double num) {
    if (number_top >= STACK_SIZE - 1)
        printf("Number Stack Overflow.\n");
    else
        number_stack[++number_top] = num;
}

// 从数字栈中取出一个数字
double pop_number() {
    if (number_top == -1) {
        printf("Number Stack Underflow.\n");
        return 0;
    } else {
        return number_stack[number_top--];
    }
}

// 将操作符入栈
void push_operator(char op) {
    if (operator_top >= STACK_SIZE - 1)
        printf("Operator Stack Overflow.\n");
    else
        operator_stack[++operator_top] = op;
}

// 从操作符栈中取出一个操作符
char pop_operator() {
    if (operator_top == -1) {
        printf("Operator Stack Underflow.\n");
        return '\0';
    } else {
        return operator_stack[operator_top--];
    }
}

// 判断是否为有效的操作符
int is_operator(char op) {
    return op == '+' || op == '-' || op == '*' || op == '/' || op == '(' || op == ')';
}

// 判断是否为数字或小数点
int is_number(char c) {
    return isdigit(c) || c == '.';
}

// 计算表达式
double calculate(double num1, double num2, char op) {
    switch(op) {
        case '+':
            return num1 + num2;
        case '-':
            return num1 - num2;
        case '*':
            return num1 * num2;
        case '/':
            return num1 / num2;
        default:
            printf("Invalid Operator.\n");
            return 0;
    }
}

double evaluate(char* expression) {
    // 初始化操作符栈和数字栈
    operator_top = -1;
    number_top = -1;
    
    int i = 0;
    double result = 0;

    while (expression[i] != '\0') {
        // 跳过空白字符
        if (expression[i] == ' ')
            i++;

        // 如果是数字或小数点，入栈
        else if (is_number(expression[i])) {
            double num = 0.0;
            int flag = 0; // 标记是否发现小数点
            while (is_number(expression[i]) || (!flag && expression[i] == '.')) {
                push_number(expression[i]);
                if (expression[i] == '.')
                    flag = 1;
                else
                    num = num * 10 + (expression[i] - '0');
                i++;
            }
            if (flag)
                num /= 10.0;
            push_number(num);
        }

        // 如果是操作符，入栈
        else if (is_operator(expression[i])) {
            if (expression[i] == '(')
                push_operator(expression[i]);
            else if (expression[i] == ')') {
                while (operator_stack[operator_top] != '(') {
                    double num2 = pop_number();
                    double num1 = pop_number();
                    char op = pop_operator();
                    push_number(calculate(num1, num2, op));
                }
                pop_operator(); // 弹出左括号
            } else {
                int current_priority = priority(expression[i]);
                while (operator_top >= 0 && priority(operator_stack[operator_top]) >= current_priority) {
                    double num2 = pop_number();
                    double num1 = pop_number();
                    char op = pop_operator();
                    push_number(calculate(num1, num2, op));
                }
                push_operator(expression[i]);
            }
            i++;
        }

        // 其他非法字符
        else {
            printf("Invalid Character.\n");
            return 0;
        }
    }

    // 处理剩余的操作符
    while (operator_top >= 0) {
        double num2 = pop_number();
        double num1 = pop_number();
        char op = pop_operator();
        push_number(calculate(num1, num2, op));
    }

    // 最终结果为数字栈中剩余的唯一元素
    if (number_top != 0) {
        printf("Invalid Expression.\n");
        return 0;
    } else {
        return pop_number();
    }
}

int main() {
    char expression[100];
    printf("Please input a math expression: ");
    fgets(expression, 100, stdin);
    double result = evaluate(expression);
    printf("Result: %f\n", result);
    return 0; 
}

您可以将需要计算的数学表达式作为程序的输入，运行即可得到计算结果。希望能对您有所帮助。