当我有一个指向结构的指针时，为什么我必须使用 malloc? [英] Why do I have to use malloc when I have a pointer to a struct?

问题描述

 struct element {
     int val;
     int z;
 };
 typedef struct element ELEM;

看看这个例子:

 int main()
 {

    ELEM z;
    z =  6;
    printf("%d",z);
 }

一切正常，但如果我有一个指向结构的指针，我需要编写这样的代码:

Everything work fine , but if I have a pointer to structure I need to write the code like this:

ELEM *z;
p = (ELEM*)malloc(sizeof(ELEM)); // Without this will not work
(*p).val = 3;
p = (ELEM*)malloc(sizeof(ELEM));
printf("%d",(*p).val);

推荐答案

声明一个指针只会创建一个指针.必须有一些东西可以指向 ， 这就是 malloc 给你的.

Declaring a pointer doesn't create anything but a pointer. Gotta have something for it to point to, which is what malloc gives you.

或者，您可以在堆栈上创建结构(又名自动存储"):

Alternatively, you could have created the struct on the stack (a.k.a. "automatic storage"):

ELEM z;
ELEM *p = &z;
(*p).val = 3; // Also could be written as p->val = 3;
printf("%d",(*p).val);

顺便说一句，你的指针代码有一个错误，因为它泄漏(即丢失了)第一个分配的结构:

BTW, your pointer code has an error, in that it leaks (i.e. loses track of) the first allocated struct:

ELEM *p;
p = (ELEM*)malloc(sizeof(ELEM));
(*p).val = 3;
p = (ELEM*)malloc(sizeof(ELEM)); // <- leak here: pointer to old struct lost.
printf("%d",(*p).val);

删除第二个 malloc 可以解决问题.一个完整的、修复过的版本，看起来更像您在使用中看到的代码:

Deleting the second malloc fixes the problem. A full, fixed-up version that looks more like code you'd see in use:

ELEM *p = (ELEM*)malloc(sizeof(ELEM));
p->val = 3;
printf("%d\n", p->val);
free(p);

每个 malloc 都应该有一个 free，除非你的程序通过终止来释放它的内存.即便如此，拥有 free 也很好.

Every malloc should have a free, unless your program releases it memory by terminating. And even then, it's nice to have the free.

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