C++ 动态分配的字符

dynamically allocated char

提问by Johannes Schaub - litb

int main()

{
   char *second= new char("hello");
   char *first="hi";
   char third[]="new";
}

I am new to c++ and don't really understand how char works, why the first one give a compiler error, and what are the differences of those 3 way of declaration , the strength and benefits of declaring it in a particular way.

我是 C++ 新手，并不真正了解 char 是如何工作的，为什么第一个会出现编译器错误，以及这 3 种声明方式的区别，以特定方式声明它的强度和好处。

Thanks

谢谢

Hmm, as someone mentions that the second form is read only, why could I Change it. suppose I have the code below

嗯，正如有人提到第二种形式是只读的，我为什么要改变它。假设我有下面的代码

 int main()

{
   char *second= new char("hello");
   char *first="hi";
   char third[]="new";
   first="world";
}

code like above will still execute, why is it so? , then which form is better if I want to read an input but don't know the size of the string?

上面的代码仍然会执行，为什么会这样？，那么如果我想读取输入但不知道字符串的大小，哪种形式更好？

回答by Johannes Schaub - litb

Know that

我知道

"abc"

allocates static storage somewhere, which lasts the whole program lifetime. You cannot write to that storage, so C++ gives it the type char const[N](an array of N constant characters). Now, the following makes a pointer point to that storage

在某个地方分配静态存储，这会持续整个程序生命周期。您无法写入该存储，因此 C++ 为其提供类型char const[N]（N 个常量字符的数组）。现在，以下内容使指针指向该存储

char *first = "hi";

Since that drops a const, that way of initializing the pointer is deprecated. That it works at all is just to keep backward compatibility with C, where a string literal does not have a const type (but is still read only). Prefer the following instead

由于这会删除 a const，因此不推荐使用这种初始化指针的方式。它完全有效只是为了保持与 C 的向后兼容性，其中字符串文字没有 const 类型（但仍然是只读的）。更喜欢以下内容

char const *first = "hi";

In constrast, the last way you have shown copies the string literal's content to an array, which will be writable, and be sized so the string literal just fits into it.

相比之下，您所展示的最后一种方式是将字符串文字的内容复制到一个数组中，该数组将是可写的，并调整大小以便字符串文字正好适合它。

char third[] = "new";

If you do that in a function, then as all variables, that array will be cleaned up when you leave its scope. Now, the first way you have shown is different. It creates a character dynamically. You could have initialized it like this

如果您在函数中执行此操作，那么作为所有变量，当您离开其作用域时，该数组将被清除。现在，您展示的第一种方式是不同的。它动态地创建一个角色。你可以像这样初始化它

char *c = new char('A');

And since that happens dynamically, you need to tell the compiler explicitly when it should free the memory

由于这是动态发生的，您需要明确告诉编译器何时应该释放内存

delete c;

But you cannot initialize the character with a string literal. What you probably had in mind is creating storage dynamically, initialized with the string literal. That's not possible using new. The only form of initializing a dynamic array is to zero it out, but you cannot directly-initialize with the content of a string literal or another array. For this form of using new, there is rarely a need to do that directly. If you want, you can do it by creating a dynamic array of the right size, and then copy bytes from the string literal to that buffer

但是您不能用字符串文字初始化字符。您可能想到的是动态创建存储，并使用字符串文字进行初始化。使用new. 初始化动态数组的唯一形式是将其清零，但您不能直接使用字符串文字或其他数组的内容进行初始化。对于这种形式的 using new，很少需要直接执行此操作。如果需要，您可以创建一个大小合适的动态数组，然后将字节从字符串文字复制到该缓冲区

char *c = new char[sizeof "hello"]; // sizeof "hello" will give 6
std::strcpy(c, "hello");

delete[] c; // delete[] is for deleting dynamic arrays

Remember that this is quite low-level, and i recommend you to use strings

请记住，这是相当低级的，我建议您使用字符串

std::string s = "hello"; // s.size() gives you its size

It completely manages memory for you. Concatenation, indexing and that stuff is available too.

它完全为您管理内存。连接、索引和那些东西也可用。

回答by elcuco

Lets try to explain in code:

让我们试着用代码来解释：

// your code
char *first="hi";
// this is the memory location of the string, assigned by the compiler
// sizeof(first) == sizeof(char*) == 4 (usually, lets not get into this right now)
char *first = 0x12345;


// your code
char third[]="new";
// means, the following:
char third[4];
third[0] = 0x6E; // ascii code of 'n'
third[1] = 0x65; // ascii code of 'e'
third[2] = 0x77; // ascii code of 'w'
third[3] = 0x00; // strings in C end with NULL
// note that sizeof(third) is still sizeof(char*), but this 
// time you statically allocated sizeof(char)*4 for the whole array

More reading for you:

为您提供更多阅读：

• http://en.wikipedia.org/wiki/ASCII
• http://www.cprogramming.com/tutorial/lesson9.html

• http://en.wikipedia.org/wiki/ASCII
• http://www.cprogramming.com/tutorial/lesson9.html