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What happens if you free a pointer twice?

If you free a pointer, use it to allocate memory again, and free it again, of course it's safe

If you free a pointer, the memory you freed might be reallocated. If that happens, you might get that pointer
back. In this case, freeing the pointer twice is OK, but only because you've been lucky. The following example
is silly, but safe:

#include <stdlib.h>
int main(int argc, char** argv)
        char** new_argv1;
        char** new_argv2;
        new_argv1 = calloc(argc+1, sizeof(char*));
        free(new_argv1);    /* freed once */
        new_argv2 = (char**) calloc(argc+1, sizeof(char*));
        if (new_argv1 == new_argv2) 
                /* new_argv1 accidentally points to freeable memory */
                free(new_argv1);    /* freed twice */
        new_argv1 = calloc(argc+1, sizeof(char*));
        free(new_argv1);    /* freed once again */
        return 0;

In the preceding program, new_argv1 is pointed to a chunk of memory big enough to copy the argv array,
which is immediately freed. Then a chunk the same size is allocated, and its address is assigned to new_argv2.
Because the first chunk was available again, calloc might have returned it again; in that case, new_argv1 and new_argv2 have the same value, and it doesn't matter which variable you use. (Remember, it's the pointed-
to memory that's freed, not the pointer variable.) new_argv1 is pointed to allocated memory
again, which is again freed. You can free a pointer as many times as you want; it's the memory you have to
be careful about.

What if you free allocated memory, don't get it allocated back to you, and then free it again? Something like

void caller( ... )
        void *p;
        /* ... */
        callee( p );
        free( p );
void callee( void* p )
        /* ... */
        free( p );

In this example, the  caller() function is passing  p to the  callee() function and then freeing  p.
Unfortunately, callee() is also freeing p. Thus, the memory that p points to is being freed twice. The ANSI/
ISO C standard says this is undefined. Anything can happen. Usually, something very bad happens.

The memory allocation and deallocation functions could be written to keep track of what has been used and
what has been freed. Typically, they aren't. If you free() a pointer, the pointed-to memory is assumed to
have been allocated by malloc() or calloc() but not deallocated since then. free() calculates how big that
chunk of memory was and updates the data structures in the memory "arena." Even if the
memory has been freed already, free() will assume that it wasn't, and it will blindly update the arena. This
action is much faster than it would have been if free() had checked to see whether the pointer was OK to

If something doesn't work right, your program is now in trouble. When free() updates the arena, it will
probably write some information in a wrong place. You now have the fun of dealing with a wild pointer;

Do you know that?

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