当结构包含字符串时为结构分配内存 [英] allocating memory for a struct when it includes a string

问题描述

假设我有这样的结构

struct node{ // note that i have changed the struct code according to my convenience

    char* lastname;
    char* employeeID;

    struct node* nextRecord;
} ;
typedef struct node myRecord;

现在我为如下所示的sirst节点分配内存

Now I am allocating memory for the sirst node as below

headptr=malloc(sizeof(myRecord));

该结构包含两个字符串.当我将某些东西存储在headptr-> lastname中时，该文件存储在哪里?我应该为这两个字符串显式分配内存吗?

The struct inludes two strings. when I store something in headptr->lastname where does it get stored? shoul I allocate memory for those two strings explicitly?

推荐答案

when I store something in myRecord->lastname where does it get stored?

这将导致 不确定的行为 .

It will lead to undefined behaviour.

should I allocate memory for those two strings explicitly?

是的，您还必须为结构成员lastname和employeeIDas分配.

Yes, you have to allocate for the struct members lastname and employeeIDas too.

赞:

headptr=malloc(sizeof(myRecord));

headptr->lastname = malloc(n1); // Alllocate n1 bytes
headptr->employeeIDas = malloc(n2); // Alllocate n2 bytes

但是，如果您为这些指针分配字符串文字，则无需分配内存.

However, if you assign string literals to those pointers, then you don't need to allocate memory.

headptr->lastname = "last name";
headptr->employeeIDas = "12345";

在这里，您正在使指针指向具有静态存储持续时间的字符串文字.

Here, you are making the pointers to point to string literals which have static storage duration.

不能在C中修改字符串文字(尝试修改会调用未定义的行为).如果要修改它们，则应该采用以前的方法(分配内存)并复制字符串文字.

String literals can't be modified in C (attempting to modify invokes undefined behaviour). If you intend to modify them, then you should take former approach (allocate memory) and copy the string literals.

headptr->lastname = malloc(n1); // Alllocate n1 bytes
headptr->employeeIDas = malloc(n2); // Alllocate n2 bytes

然后复制它们:

strncpy(headptr->lastname, "last name", n1);
headptr->lastname[ n1 - 1 ] = 0;
strncpy(headptr->employeeIDas, "12345", n2);
headptr->employeeIDas[ n2 - 1 ] = 0;

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