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Nov 29th, 2005

Memory allocation for typedef struct

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Hello:
I have a question regarding struct when making hash buckets eg.
typedef struct _hashNode
{
void *key;
void *value;
struct _hashNode *next;
} hashNode;

let's say when I create the table:

hashNode *nodeList = (hashNode *)malloc(sizeof(hashNode) * 200);

This is because I want a fixed list size and then each of these array spots will be a linked list or "buckets" for inserting values downwards. (I hope you understand this) Anyways, if I were to insert values into lets say nodeList[5]->key, would I have to allocate memory to key

eg. nodeList[5]->key = (void *)malloc(sizeof(void) +1);

Or does that automatically do that when I allocate memory to the nodeList[5].

thanks

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desertstorm

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Nov 29th, 2005

Re: Memory allocation for typedef struct

Quote originally posted by desertstorm ...

Or does that automatically do that when I allocate memory to the nodeList[5].

No, you need to do that.

[edit]
Except that this won't go far: sizeof(void) -- you need a type to have a size of that type.
[/edit]

Why the casts on malloc? If you're using C++, why not new or a vector? If C, stop doing that and let your compiler help you write better code.

Dave Sinkula

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Nov 29th, 2005

Re: Memory allocation for typedef struct

The purpose of malloc is dynamically, key word, dynamically allocate memory from the preprocessor for a DATA structure.

In this case hashNode or whatever is your data structure, so you would be mallocing a pointer to an array of hashNode structures. Each of these hashNodes have the elements you specified under the structure definition. They each have a keys, and so forth, without further mallocing.

Now if you wanted a malloc'd array of hashNodes and a malloc array of keys, that would be a little difficult to handle with, but I guess it could be possible, althought it is definately poor code.

malloc is a data structure preprocessor function, keep this in mind. It will not change or modify the elements in the data structure, it simply allocates memory.

Hope this helps,

Mistro116

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Nov 29th, 2005

Re: Memory allocation for typedef struct

The thing is I am setting node[5]->key for example to another pointer eg. void *key. So something like this: node[5]->key = key. I am setting the pointer in the typedef struct to the pointer void *key which is an argument/parameter in the function header.

eg. void *insert(void *key)
{
node[5]->key = key;
}

node is global.

If done this way I don't have to allocate memory to node[5]->key before setting it equal to key right?

desertstorm

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Re: Memory allocation for typedef struct

C++ Syntax (Toggle Plain Text)
 node[5]->key = key;
 node[5]->key = key;

That is wrong syntax. you have to use dot notation, not pointer notation because node[5] is NOT a pointer.

C++ Syntax (Toggle Plain Text)
 node[5].key = key;
 node[5].key = key;

Quote ...

If done this way I don't have to allocate memory to node[5]->key before setting it equal to key right?

Well, maybe yes and maybe no. How is paramater key being allocated? Will the calling function reuse that memory for something else, such as does it get overwritten each time the insert function is called or is new memory allocated each time?

I find it normally best to reallocate the memory so that node has complete control over it and not count on some other function allocating/deallocating node's memory.

C++ Syntax (Toggle Plain Text)
eg. void *insert(void *key, long keysize)
{
node[5].key = malloc(keysize);
memcpy(node[5].key,key,keysize);
}
eg. void *insert(void *key, long keysize)
{
node[5].key = malloc(keysize);
memcpy(node[5].key,key,keysize);
}

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