In this code snippet ,realloc is equivalent to malloc but I am unable to get the logic .
int *ptr=(int*) realloc(NULL,10*sizeof(int));
why does it creates a new block , since NULL is a macro defined in stdio.h as 0 so it implies that it points to the base address 0 which in most machines is system area , so how can realloc start allocating a memory from base address 0 for 10 integers , why is this not a segmentation fault ?
To clarify: it means the function realloc checks if its first argument is 0 (a NULL pointer) and if yes, behaves like malloc or simply calls malloc.
so how can realloc start allocating a memory from base address 0
From realloc manual:
In case that ptr is a null pointer, the function behaves like malloc, assigning a new block of size bytes and returning a pointer to its beginning.
So in case the previous pointer is NULL, it doesn't mean that realloc has to start allocating from base address 0. In fact, it will behave like malloc, allocating a new block of memory.
NULL is a macro defined in stdio.h as 0 so it implies that it points to the base address 0
That's not true. NULL is the macro for the null pointer, which doesn't necessarily point to base address 0. See C FAQ.
To answer your main question, realloc behaves like malloc when the first argument is the null pointer.
Yet another answer, All have explained beautifully but misses some good observation.
The following behaviour is taken from Linux man. You might like read doc for your platform. realloc(void *ptr, size_t size) has been to have a bad design, trying to perform many task, for instance
1) if ptr == NULL, it behaves like malloc(size)
2) if ptr != NULL && size == 0, it behaves like free(ptr)
More details at man 3 realloc.
So, it is preferable to have a wrapper for realloc especially for usage
like ptr = realloc(ptr, size); which leads to leak.
void* my_realloc(void *oldPtr, size_t newSize)
{
void *newPtr = NULL;
if ( (newPtr = realloc( oldPtr, newSize) ) == NULL)
{
free( oldPtr);
return NULL;
}
return newPtr;
}
realloc(some_pointer, 0) returns NULL or a non-NULL pointer. IMO, a weakness in realloc() specification. my_realloc() does the same - no improvement there. Better if realloc(some_pointer, 0) returned non-NULL so a return value of NULL only occured with an error.realloc(not_null_pointer, 0) may return a valid (not-NULL) pointer. It is not like free().