Re: Unable to free malloc'd ptr
"Jason Pawloski" <jpawloski@gmail.com> wrote in message
news:1182379285.195245.4720@p77g2000hsh.googlegroups.com...
On Jun 20, 3:19 pm, Ian Collins <ian-n...@hotmail.com> wrote:
Jason Pawloski wrote:
Relevant:
I have a local integer pointer:
int *nr1 = NULL;
within this function, I call another function
int *function2([parameters])
This function mallocs an int *ptr and returns it. So I use the
function like:
nr1 = function2([parameter]);
Now when I try to free it later in my first function (the one that
calls function2), I get an MSVC++ error. The exact wording is:
Debug Error!
Program: <path>
DAMAGE: after Normal block (#46) at 0x00431D80
Any suggestions on what might cause this? thanks!
Can you post function2?
--
Ian Collins.- Hide quoted text -
- Show quoted text -
Sure. Again, not a programmer by trade, etc.:
NB size is the absolute length of *list (ie number of elements), while
position is the INDEX of what to remove from the list. Confusingly
stupid, I know, but working it this way makes other things in the
program not relevant to this simple.
int *dropreel(int *list,int size,int position)
{
int *nl = NULL;
/*Put the index in range */
while(position > size-1) position = position - size;
while(position < 0) position = position + size;
nl = malloc(sizeof(int)*(size-1));
if(nl == NULL)
printf("WARNING!! NO MEMORY!\n");
memset(nl,0,sizeof(int)*(size-1));
if(position != size-1)
nl[0] = list[size-1];
else
nl[0] = list[size-2];
memcpy(&nl[1],list,sizeof(int)*(position));
memcpy(&nl[position+1],&list[(position+1)],sizeof(int)*(size-
position-1));
return nl;
}
It's a little confusing code, but it seems that your list[0] stores the size
of the list, with list[1] starting the actual data. Does the size parm
being passed in include one for list[0] ? That is, are you simply passing
size as list[0] into the function?
And if so, why are you decrementing size in the malloc? For example, if you
had a list of three values:
list[0] = 3;
list[1] = 1;
list[2] = 2;
list[3] = 3;
you would be passing in size as 3, although it actually contains 4 ints.
Now, say you are wanting to remove element number 2. So you would pass
list, list[0], 2
correct?
Now, when you malloc, sizeof(int) * (size - 1)
size is 3. 3-1 = 2. You are only allocating 2 ints, when you acutally need
3.
This is obvoiusly C code and not C++ (the giveaway is malloc's return value
not being cast which is required in C++) otherewise I'd say just use
std::vector and be done with it.
Your while code is somewhat flakey (while (position > size - 1, etc...) and
if it was me, if position was > size I'd print an error and return null.
It sounds most likely like you are not allocating enough memory and
overflowing the array with your memcpys.