释放calloc，结构与C指针 [英] Calloc with structure with pointers in C

问题描述

我知道要使用的释放calloc申请的内存，对所有位写0，然后返回一个指针。

I know that calloc request memory to be used, writes 0 on all the bits and then returns a pointer to it.

我的问题是：如果我使用释放calloc与包含指针的结构，将这些指针有NULL值，或者我必须将它们设置为指向NULL

My question is: if I use calloc with a structure that contains pointers, will those pointers have the NULL value or do I have to set them to point to NULL?

struct a{
char * name;
void * p;
}* A;

所以基本上，我用释放calloc与结构后，将其命名为和P点为NULL？

So basically, will name and p point to NULL after I've used calloc with struct a?

谢谢！

推荐答案

不知怎的，你已经得到了很多不正确的答案。 C不要求空指针重新presentation是所有零位。很多人误以为这样做，因为一个整型常量前pression值为0，转换为指针，成为一个空指针。

Somehow you've gotten a lot of incorrect answers. C does not require the representation of null pointers to be all-zero-bits. Many people mistakenly think it does, since an integer constant expression with value 0, converted to a pointer, becomes a null pointer.

随着中说，在所有现实世界的系统中，空指针都是零位，而释放calloc 是一个完全合理的方式来获得一个空指针 - 初始化指针阵列中的现实世界

With that said, on all real-world systems, null pointers are all-zero-bits, and calloc is a perfectly reasonable way to get a null-pointer-initialized pointer array in the real world.

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