在c ++中声明变量和动态分配内存到变量？ [英] Declaring a variable vs dynamically assigning memory to a variable in c++?

问题描述

我有点困惑，当我只是声明一个变量，如：

I'm a bit confused about the difference between when I just declare a variable such as:

int n;

并使用new将内存动态分配给变量，例如：

and dynamically assigning memory to a variable using "new" such as:

int m = new int;

我注意到只是从一个简单的链表项目，当我插入一个新的值节点对象的形式，我必须动态创建一个新的节点对象，并附加所需的值到它，然后链接到我的列表的其余部分。但在同一个函数中，我可以只定义另一个节点对象，ex。 NodeType * N。并使用此指针遍历我的列表。
我的问题是..当我们只是声明一个变量，内存不会立即分配..或有什么区别？

I noticed just from working on a simple linked list project that when I'm inserting a new value in the form of an node object, I have to dynamically create a new node object and append the desired value to it and then link it to the rest of my list. However.. in the same function, I could just define another node object, ex. NodeType *N. and traverse my list using this pointer. My question is.. when we just declare a variable, does memory not get assigned right away.. or what's the difference?

谢谢！ p>

Thank you!

推荐答案

如有可能，首选自动存储分配的变量：

Prefer automatic storage allocated variables when possible:

int n;

过

int* m = new int; // note pointer

动态分配的首要原因是链表定义的方式。也就是说每个节点包含指向下一个节点的指针（可能）。因为节点必须存在，超出它们创建的点，所以它们是动态分配的。

The reason dynamic allocation is prefered in your case is the way the linked list is defined. I.e. each node contains a pointer to a next node (probably). Because the nodes must exist beyond the point where they are created, they are dynamically allocated.

NodeType * N。并使用此指针遍历我的列表

NodeType *N. and traverse my list using this pointer

是的，你可以这样做。但请注意，这只是一个指针声明。

Yes, you could do that. But note that this is just a pointer declaration. You have to assign it to something meaningful to actually use it.

我的问题是..当我们只声明一个变量，是否记忆不是

My question is.. when we just declare a variable, does memory not get assigned right away.. or what's the difference?

实际上，这两种情况都是定义，而不仅仅是声明。

Actually, both cases are definitions, not just declarations.

int n;

创建一个未初始化的 int 存储;

creates an un-initialized int with automatic storage;

int* n;

创建指向 int 的指针。它是悬空的，它不指向有效的内存位置。

creates a pointer to an int. It's dangling, it doesn't point to a valid memory location.

int* n = new int;

创建一个指针并将其初始化到包含未初始化 int的有效内存位置。

creates a pointer and initializes it to a valid memory location containing an uninitialized int.

int* n = new int();

创建一个指针并将其初始化到一个有效的内存位置，包含一个值初始化的 int （即 0 ）。

creates a pointer and initializes it to a valid memory location containing a value-initialized int (i.e. 0).

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