双指针上的memset操作

Question

Question is relevant.

For the below representation,

  typedef struct List{

    void **array; // array of void*
    int lastItemPosition;
    int size;
  }List;

  #define INITIAL_LIST_SIZE 50

createList performs as shown below,

List *createList(List *list, Op opType){

  List *lptr = (List *)malloc(sizeof(List));

  if(opType == CREATE_NEW_LIST){

    lptr->array = malloc(INITIAL_LIST_SIZE*sizeof(void*));
    lptr->array = memset(lptr->array, NULL, INITIAL_LIST_SIZE*sizeof(void *));
    lptr->lastItemPosition = -1;
    lptr->size = INITIAL_LIST_SIZE;
}

Is memset performing valid operation on lptr->array ?

Answer 1

In your code,

 memset(lptr->array, NULL, INITIAL_LIST_SIZE*sizeof(void *));

is wrong, as the second argument is expected to be an int , you're passing a pointer. The conversion is a highly implementation defined behaviour, and in most of the cases, it would invoke UB.

Related, quoting C11 , chapter §7.19

NULL
which expands to an implementation-defined null pointer constant; [...]

and, chapter §6.3.2.3

An integer constant expression with the value 0, or such an expression cast to type void * , is called a null pointer constant .

So, NULL is of pointer type which is not the compatible type to an int in any ways.

Answer 2

It's valid on all major platforms, except for one thing: Don't pass NULL as the value to set. Remember that the memset function operates on the individual bytes of the memory, and you should set all the bytes to zero ( 0 ).

It is however not strictly technically valid. On most major platforms a null pointer is equal to zero. But it doesn't have to be that. The only fully portable and safe way to do it is through a manual loop where you set each pointer explicitly to NULL .

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And if anyone is interested to know, even if NULL is defined as 0 (or ((void *) 0) ) it doesn't matter. The compiler will translate that zero into the platform-specific version of a null pointer, which may be something else than the actual integer zero.