指针和字符串C++

我在教自己C++,我对指针有点困惑(特别是在下面的源代码中).但首先,我继续向你展示我所知道的内容(然后将代码与此对比,因为我觉得好像存在一些矛盾).

我知道的:

int Age = 30;
int* pointer = &Age;

cout << "The location of variable Age is: " << pointer << endl;
cout << "The value stored in this location is: " << *pointer << endl;

指针保存内存地址.使用间接(取消引用)运算符(*),可以访问存储在指针的内存位置的内容.在本书的代码中,我无法理解......

cout << "Enter your name: ";
string name;
getline(cin, name); //gets full line up to NULL terminating character

int CharsToAllocate = name.length() + 1; //calculates length of string input
                          //adds one onto it to adjust for NULL character
char* CopyOfName = new char[CharsToAllocate];
// pointer to char's called CopyOfName, is given the memory address of the 
//beginning of a block
//of memory enough to fit CharsToAllocate. Why we added 1? Because char's need a 
//NULL terminating character (\0)

strcpy(CopyOfName, name.c_str()); //copies the string name, into a pointer?

cout << "Dynamically allocated buffer contains: " << CopyOfName << endl;
delete[] CopyOfName; //always delete a pointer assigned by new to prevent memory leaks

输出:

Enter your name: Adam
Dynamically allocated buffer contains: Adam

上面代码中的注释是我的评论.我的问题始于strcpy.为什么要name.c_str()复制到指针CopyOfName？这是否意味着所有字符串都是必不可少的指针？就像字符串测试="Hello world"; 实际上是指向存储"H"的存储位置的指针吗？

接下来,为什么它在print out语句中使用CopyOfName而不是*CopyOfName？指针有内存地址吗？使用*CopyOfName会打印出内存位置的内容.我在Code :: Blocks中尝试了这个,如果输入文本是"Hello World".*CopyOfName在打印输出语句中使用只会给出"H".这是有道理的,因为当我声明我需要一个带有'new'的内存块时,这实际上会返回一个指向动态分配的内存块的第一部分的指针.

我可以调和它的唯一方法是字符串实际上是一个指针.

string testing = "Confused";
cout << testing << endl;

会打印出"困惑"这个词

但是,如果我尝试编译

string testing = "Confused";
cout << *testing; 

我收到一条错误消息.

基本上,总结一下我的问题,我试图理解代码strcpy和cout声明.