Malloc - question

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Hi,
Enclosed my phonebook code.
could you please explane the need of this line:

Code:

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phonebook=(phone*)malloc(sizeof(phone));

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This line is needed but I don't understand why ...

Thanks in advanced

Code:

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int main(void)
{
    char userChoice;

    phone *phonebook;
    phone *phonebook_ptr[100];
  phonebook=(phone*)malloc(sizeof(phone));
    phonebook_ptr[counter]=(phone*)malloc(sizeof(phone));

.....

}

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malloc() creates a persistent block of memory, that lasts until you release it with free().

It's hard to show the use in a tiny fragment. Normally there's no point in mallocing a single object, then freeing it, in main().

You need malloc() 1) when you are allocating a lot of objects, and you don't know the number at compile time. 2) when you are passing data between functions a lot and need to maintain a "state" across calls.

So if I understand corectly ...

Could you please approve ?

phone *phonebook; phone *phonebook_ptr[100]; phonebook=(phone*)malloc(sizeof(phone)); // because - passing data between functions phonebook_ptr[counter]=(phone*)malloc(sizeof(phone)); // because - don't know the number at compile time

Post your code in [code]/*your code*/[/code] .

After looking through your code I think your problem is that you don't understand the difference between pointers and arrays.

If you want to have an array of "phone" elements you either need to use a static array like

Code:

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phone phonebook[100];

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or a dynamic array like

Code:

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phone *phonebook = malloc(sizeof(*phonebook) * nr_of_entries);

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Your function declarations and calls would look like:

Code:

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void PrintEntries(phone pbook[], int size);  // plural because you print all entries not just one
...
PrintEntries(phonebook, current_size);

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Using a second parameter for the actual size of the array gets rid of the global counter.

Bye, Andreas

Tanks for the answers but I don't think I understend the answers for my question.

enclosed ( again) all my codeAttachment 12587Attachment 12588Attachment 12589

Code:

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int main(void) {          phone *phonebook;    phone *phonebook_ptr[100];    phonebook=(phone*)malloc(sizeof(phone)); // WHY need malloc here ? I don't understand    phonebook_ptr[counter]=(phone*)malloc(sizeof(phone)); //since don't know the data size  .....  }

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For reference -> Malloc - warning - Dev Shed

Salem, you truly are everywhere!

Quote:

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Originally Posted by ilans11il

Tanks for the answers

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You are welcome.

Sorry ,
but I don't understend the answer
why do I need this line:

phonebook=(phone*)malloc(sizeof(phone));

thanks

Because phonebook is just a pointer. It points to nowhere.
You want this pointer to point at some memory, where data are going to be stored and processed by the programmer. Malloc allocates that memory for you and it returns a pointer to that block of memory.

Quote:

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Originally Posted by ilans11il

Sorry ,
but I don't understend the answer
why do I need this line:

phonebook=(phone*)malloc(sizeof(phone));

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The question you are asking is totally irrelevant. Your code is broken due to other reasons. The line you are asking about just keeps the compiler quiet but is the wrong solution.

You have to understand how arrays work and how to pass them to functions. Start with static (fixed sized) arrays. You can think about dynamic arrays later.

You could start with Arrays in C. Then you could continue with Arrays as Parameters and the grand final would be A Tutorial on Pointers and Arrays in C.
There is also the C-FAQ which has a whole section about Arrays and Pointers (I think questions 2, 3, 4, 8, 10, 12, 13 and 14 are interesting for you at the moment).

Bye, Andreas

Thanksssssssssss
for the explenation and the greate info - I'll use it :o