Why do I get memory leak? - memory

Question

Consider the following piece of code:
#include
#include
int main () {
char *str;
/* Initial memory allocation */
str = (char *) malloc(15);
strcpy(str, "tutorialspoint");
printf("String = %s, Address = %u\n", str, str);
str = NULL;
free(str);
return(0);
}
Why does the above program cause a memory leak? How do I avoid this?
An error is thought to occur in "str = NULL;". Why?
valgrind log:
==4143== Memcheck, a memory error detector
==4143== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==4143== Using Valgrind-3.13.0 and LibVEX; rerun with -h for copyright info
==4143== Command: ./a.out
==4143==
String = tutorialspoint, Address = 86097984
==4143==
==4143== HEAP SUMMARY:
==4143== in use at exit: 15 bytes in 1 blocks
==4143== total heap usage: 2 allocs, 1 frees, 1,039 bytes allocated
==4143==
==4143== 15 bytes in 1 blocks are definitely lost in loss record 1 of 1
==4143== at 0x4C2FB0F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==4143== by 0x1086EB: main (in /home/stack/a.out)
==4143==
==4143== LEAK SUMMARY:
==4143== definitely lost: 15 bytes in 1 blocks
==4143== indirectly lost: 0 bytes in 0 blocks
==4143== possibly lost: 0 bytes in 0 blocks
==4143== still reachable: 0 bytes in 0 blocks
==4143== suppressed: 0 bytes in 0 blocks
==4143==
==4143== For counts of detected and suppressed errors, rerun with: -v
==4143== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)

Answer 1

free(str); frees the space pointed to by str, where str is the space obtained by malloc. Because of the line str = NULL; happens before the free, free is attempting to deallocate the memory position at location 0. By definition in the C standard, this does nothing. It is common to set a pointer to 0 after it is deleted, so that if one were to accidentally attempt to delete it again, nothing happens.
To fix you code, you simply need to swap the lines str = NULL; and free(str);