Question

I ve the following memory allocation done
int *a;
a=malloc(5*sizeof(int));
Is there a way to just free say the 5th location of this set?

ie can I do free(a+4)??

Currently this gives a segmentation fault error.

Answer 1

No, you cannot do free(a+4) etc.

Why? Because the C standard forbids it.
You must pass the exact pointer returned by malloc() to free() without any pointer arithmetic.

• http://www.kernel.org/doc/man-pages/online/pages/man3/free.3.html

  The free() function frees the memory space pointed to by ptr, which must have
  been returned by a previous call to malloc(), calloc() or realloc().

  • The Linux man pages are not the C standard itself but practically covers equivalent information on the C library functions.

Why the standard forbids it? It is because some implementations of malloc() stores header information at the location a-4 (or wherever) and free() internally reads it by pointer arithmetic. If it’s given a shifted pointer like a+4, it will miss the header info and get screwed.
Cf.

• Where Do malloc() / free() Store Allocated Sizes and Addresses?

Incidentally some libc implementations provide non-standard functions which return the size of the allocated memory chunk:

• Is it possible to find the Memory Allocated to the Pointer, without searching for the malloc statement

If you’re curious how they’re implemented and look into their source code, you often find that they read the header info at a-4 etc:

• glibc malloc/malloc.c#l698
• glibc malloc/malloc.c#l1227

OK, back to the original question.
If you want to shorten (or lengthen) the memory chunk you allocated with malloc(), you can use realloc() as @Gustav said. But if you want to remove a few bytes from the middle of your memory chunk, you need manual re-allocation and copying. Actually realloc() might internally do a similar work, so be careful with the returned pointer which may be different from the original one.