我试图创建一个类似于printf（）的自定义函数，但在精确打印浮点数时遇到错误。当我试图处理格式说明符的精度部分时，例如%.2f，就会出现问题。如何在自定义的类似printf（）的函数中实现浮点数的精度处理？

**main函数：**有我想打印的文本：

int main(void)
{
    fprint("first I fixed the integer call u can see here I can print an integer %d, I also can print a character: %c and a string: %s, and lastly I can print a float value like: %.2f\n", 567, 'a', "Hey", 65.13);
    fprint("I can print this too: %%, without a problem\n");
    return (0);
}

字符串

我常用的printf（）类函数：

void fprint(char *str, ...)
{
    va_list args;
    int length = strlen(str);
    char pr = '%';
    char *s_holder, c_holder;

    va_start(args, str);

    for (int i = 0 ; i < length ; i++)
        if (str[i] != '%')
    {
        write(1, &str[i], 1);
    }
    else if (str[i] == '%')
    {
        switch (str[i+1]){
            case 'd':
                print_integer(va_arg(args, int));
                i += 1;
                break;
            case 'c':
                c_holder = va_arg(args, int);
                write(1, &c_holder, 1);
                i += 1;
                break;
            case 's':
                s_holder = va_arg(args, char *);
                write(1, s_holder, strlen(s_holder));
                i += 1;
                break;
            case 'f':
                print_double(va_arg(args, double));
                i += 1;
                break;
            case '.':
                if (str[i + 2] >= '0' && str[i + 2] <= '9')
                {
                    print_double2(va_arg(args, double), str[i + 2]);
                    i += 3;
                }
                else
                {
                    write(1, &str[i], 1);
                }
                break;
            default:
                write(1, &str[i], 1);
                break;
        }
    }
}

型

处理浮点数的函数：

void print_double2(int num, char cprecision)
{
    int precision = atoi(cprecision);

    //handling the negative numbers
    if (num < 0)
    {
        write(1, "-", 1);
        num = -num;
    }

    //separating the integer part from the fractional part
    int int_part = (int)num;
    double fract_part = num - int_part;
    
    int int_fract;
    char digit;

    //printing the integer part of the double number with the print_integer function
    print_integer(int_part);

    write(1, ".", 1);

    //printing the fraction part
    while (fract_part > 0 && precision > 0)
    {
        fract_part *= 10;
        int_fract = (int)fract_part; //getting the int part after mult by 10
        digit = '0' + (int_fract % 10);
        write(1, &digit, 1); //printing the integer part
        fract_part -= int_fract; //removing the integer part to start the process again
        precision--; //decrementing the precision until it get to 0.
    }
}

型
我真的不知道该怎么办，我还是一个初学者，我一个月前才开始学习。我试着用chatgpt，但它没有帮助。😓