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calloc() and free()

Posted on 2001-08-15
1
539 Views
Last Modified: 2012-05-04
In a project, function1 calls function2 in another file:

in file1:
void function1()
{
for (i=1; i<5; i++)
function2();
}

in file2:
float *point1;
function2()
{
point1 = (float *) calloc(number1, sizeof(float));
...
free(point1);
}

Any comment on this code and can you tell me that after free(point1), can point1 be calloc() again?
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Question by:markdot
1 Comment
 
LVL 16

Accepted Solution

by:
imladris earned 60 total points
ID: 6390296
Nope.
Yup.

There is no apparent conceptual problem with this code. It is intended that a pointer be capable of being set to point to allocated memory. Then you can free that memory. Then you can set the pointer to point to new allocated memory. The issue with lots of dynamic allocation is to make sure that there is a corresponding "free" for every "calloc". If you miss a free, then you will have a memory leak.

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