在结构内动态分配结构

问题描述

我正在动态分配一个结构不同的结构作为成员:

I am dynamically allocating a struct which has a different struct as a member:

struct a {
   // other members
   struct b;
}

struct b基本上持有指向另一个struct b的指针，因此请将struct b视为链接列表.

struct b basically holds a pointer to another struct b, so think of struct b as a linked list.

如果我动态分配struct a，那么这也会在其中分配一个新的struct b.但是，这样做还是让struct a持有指向struct b的指针并在struct a中动态分配struct b之间有什么区别?实施上有什么区别?

If I dynamically allocate struct a, then that would also make a new struct b within it. However, what is the difference between doing that or having struct a hold a pointer to struct b, and dynamically allocate struct b within struct a? What is the difference in implementation?

推荐答案

如果您像这样动态分配(malloc)struct a

If you dynamically allocate (malloc) struct a as in

struct a *temp = (struct a *)malloc(sizeof(struct a));

您在malloc空间中放置了指向struct b的指针(假设这是struct a中的内容)，但是您没有在malloc空间中使用struct b.这意味着以后您将不得不做

you malloc space for a pointer to struct b (assuming that's what is in struct a) but you don't malloc space for struct b. That means later you'll have to do

temp->b = (struct b *)malloc(sizeof(struct b));

在尝试使用struct b之前.

如果不存储指向struct b的指针，而是直接存储struct b的指针，则在定义struct a时将获得自动分配.

If you don't store a pointer to struct b but rather struct b directly then you'll get the automatic allocation when you define struct a.

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