Re: string in an array
- From: gordonb.d55ud@xxxxxxxxxxx (Gordon Burditt)
- Date: Tue, 31 Mar 2009 17:42:40 -0500 (CDT)
My doubt comes when declaring a array bigger than the string it is
going to be stored inside it:
char* array = malloc( sizeof(char)*1024 );
The variable named 'array' is a *POINTER*, not an array.
strcpy( array, "hello" );
When comes to free it, I always supposed that the entire array is
freed.
But today, using the (non compliant) mode to check the size of an
array:
sizeof(array) / sizeof(array[0])
This doesn't work when applied to a (char *) pointer. You get the
size of a (char *) pointer. On what machine is the size of a pointer
equal to sizeof("hello")?
I noticed that the size I get is the size of "hello" and not 1024
bytes (I repeat, I know this is not the best way to check an array's
size). I'm not sure if free() is going to free strlen("hello") bytes
or the full 1024.
It works fine on an array, but you are not giving it an array.
The way to keep track of the size of a piece of memory allocated
with malloc() or friends calloc() or realloc() is to remember how
much you allocated in the first place.
This doubts are related to deleting (freeing) the source string I pass
to a function like strtok(), which is going to modify it.
Could someone clear my mind?
free() frees memory allocated by one call to malloc() or friends,
regardless of the contents of the memory. It is not possible to
free() *part of* the memory allocated by one call to malloc() or
friends.
--
comp.lang.c.moderated - moderation address: clcm@xxxxxxxxxxxx -- you must
have an appropriate newsgroups line in your header for your mail to be seen,
or the newsgroup name in square brackets in the subject line. Sorry.
.
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