Deallocating a character pointer array

I'm a noob to C++ so please excuse my ignorance.

I declared a character pointer array as follows and initialized it.

Code:

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char* c[30];

for(i=0; i<30; i++)
  c[i] = new char[20];

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I tinker around with it for a while, then I try to deallocate its memory.

Code:

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delete [] c;

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Somehow, this would cause an assertion failure (I'm using VC++ 2003).

So I tried the following

Code:

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delete [] *c;

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There was no error with this one, but it only ended up deallocating the first element of the array, and none of the other 29 elements were affected.

Is there another way to free up memory for character arrays (or do I have to iterate through each element and delete it)?

One final question. I initialized each char pointer to have 20 characters. But when I check on it in runtime, it would always have 80 characters (with one or two exceptions).

I tried switching the numbers to 5, 10, 30, whatever, and the length of the character array is always 80. Is there any way I can determine this behavior?

Thanks.

Code:

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delete [] c;

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That won't work. You need to free it in a for loop:

Code:

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for(x = 0; x < 30; x ++)
    delete [] c[x];

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[edit]

Code:

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delete [] *c;

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is the same as

Code:

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delete [] c[0];

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[/edit]

Do exactly the same, only in reverse

for(i=0; i<30; i++)
c[i] = new char[20];


for(i=0; i<30; i++)
delete [] c[i];

Beat you. :)

Quote:

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... when I check on it in runtime, it would always have 80 characters (with one or two exceptions).

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Are you sure? You can't easily check the size of the allocated memory.

[edit]

Quote:

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Is there another way to free up memory for character arrays (or do I have to iterate through each element and delete it)?

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That's how you allocated it. That's how you have to free it.
[/edit]

I don't know if this is the same as checking the size of allocated memory, but I tried printing out the following in my code:

Code:

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printf("The address offset between %p and %p is %d", c[i], c[i+1], c[i+1]-c[i]);

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And the offset was usually 80 (with one exception of 3000 something)

> And the offset was usually 80 (with one exception of 3000 something)
You can't compare the two results from two separate new() calls and get any kind of meaningful (ie portable) answer.

That just prints the difference between the addresses of the allocated pointers. They aren't guarenteed to be contiguous. (In fact, they usually aren't.) You can assume that new allocated whatever size you passed it (unless new returns NULL).

Okay, got that. Off to the next task, then.

Thanks a lot, folks.

Quote:

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Originally Posted by cunnus88

I don't know if this is the same as checking the size of allocated memory, but I tried printing out the following in my code:

Code:

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printf("The address offset between %p and %p is %d", c[i], c[i+1], c[i+1]-c[i]);

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And the offset was usually 80 (with one exception of 3000 something)

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Try this:

printf("The address offset between %p and %p is %d", &c[j][0], &c[j][29], &c[j][29] - &c[j][0]);