c++ - 如何在构造函数中初始化char

Question

    #include <iostream>
    using namespace std;

    class MyClass
    {
        private :
        char str[848];

        public :

        MyClass()
        {

        }

        MyClass(char a[])  
        {
            str[848] = a[848];
        }

        MyClass operator () (char a[])
        {
            str[848] = a[848];
        }

        void myFunction(MyClass m)
        {

        }

        void display()
        {
            cout << str[848];
        }
    };

    int main()
    {   
        MyClass m1;  //MyClass has just one data member i.e. character array named str of size X
                                //where X is a constant integer and have value equal to your last 3 digit of arid number
        MyClass m2("COVID-19") , m3("Mid2020");
        m2.display(); //will display COVID-19
        cout<<endl;
        m2.myFunction(m3);
        m2.display(); //now it will display Mid2020
        cout<<endl;
        m3.display(); //now it will display COVID-19
      //if your array size is even then you will add myEvenFn() in class with empty body else add myOddFn()
      return 0;    

    } 

我不能使用 string，因为有人告诉我不要这样做，因此，我需要知道如何制作它才能显示所需的输出

Answer 1

如何在构造函数中初始化char数组？

使用循环逐元素复制元素:

MyClass(char a[])  
{
    //make sure that sizeof(a) <= to sizeof(str);
    // you can not do sizeof(a) here, because it is
    // not an array, it has been decayed to a pointer

    for (int i = 0; i < sizeof(str); ++i) {
        str[i] = a[i];
    }
}

使用std::copy中的<algorithm>

const int size = 848;
std::copy(a, a + size, str); 

如果需要使用 std::copy，则首选 strcpy而不是 strcpy，而不是 strncpy。您可以为其提供大小，以帮助防止错误和缓冲区溢出。

MyClass(char a[])  
{
    strncpy(str, a, sizeof(str));
}

使用库中的std::array。它具有多种优势，例如，您可以像普通变量一样直接分配它。示例:

std::array<char, 848> str = {/*some data*/};
std::array<char, 848> str1;
str1 = str;