为什么两个char指针之间的字符串复制不起作用？

Question

I am trying to copy a string from one char * to another, and do not know why the copy does not work.

I am writing a linked list program -- Linklist -- and there are two char * pointers involved. Each points to a struct Node as follows:

struct Node
{
    char * message;
    char * text;
    struct Node * next;
};

typedef struct Node * Linklist;

I have written a function which has two parameters to create a new LinkNode :

Linklist create(char *message,char * text)
{
    Linklist list =(Linklist)malloc(sizeof(struct Node));
    //the message changes after the sentence but text is right.
    if(list==NULL) printf("error:malloc"); 
    list->message=message;
    list->text=text;
    return list;
}

In main:

char *message is"helloworld"

char *text is"test"

I watched the message in gdb,after malloc. The message changed to "/21F/002", but text is still "test"

I added const before message, but it doesn't work.

Can anyone can tell what is happening?

Thanks.

Answer 1

The problem is that strings in c don't work the same way. Here is how you copy a string:

Linklist create(char *message,char * text)
{
    Linklist list =(Linklist)malloc(sizeof(struct Node));
    //the message changes after the sentence but text is right.
    if(list==NULL) printf("error:malloc"); 

    list->message = malloc(strlen(message)+1);
    if(list->message==NULL) printf("error:malloc"); 
    strcpy(list->message,message);

    list->text = malloc(strlen(text)+1);
    if(list->text==NULL) printf("error:malloc"); 
    strcpy(list->text,text);

    return list;
}

Of course you have to be careful here, make sure message and text are not coming from the user or you risk buffer overflow vulnerabilities.

You can use strncpy() to solve that issue.

Answer 2

您必须为指针消息和文本分配存储空间，然后复制字符串。