函数中的双指针解引用

Question

I'm trying to understand double pointers. I think i should be able to write (**min = 3;) in the MinMax loop below, but it is just ignored. I think it should set the value pointed to by the value of min (pointer to start) to 3 (or of course to any int). Can someone help me understand why this is crazy talk? There is no reason in the function to set this value of course, i just want to understand why it doesn't work.

int ar[] = {1,23,4,32,5,67,999,-1};
int *min= 0;
int *max= 0;
MinMax(ar,ar+8,&min,&max);

void MinMax(int *start,int *end, int **min,int **max) {
  // **min = 3; //why not?
  *min = start; 
  *max = start;
  while(++start < end) {
  if(*start < **min) *min = start;
  if(*start > **max) *max = start;
  }
}

Answer 1

At the point you have **min = 3 the value of *min is 0, ie a NULL pointer, which means it doesn't point anywhere. Attempting to dereference *min and subsequently write to it invokes undefined behavior .

The following lines set both *min and *max to point to the same place as start , so after that they can be dereferenced.

Answer 2

On entering the function *min is zero so **min = 3 is setting an integer at address zero to three. Not normally allowed at run time. It is only after initialising *min to a valid address, for example *min = start, that you can then set **min to a value.

Answer 3

void MinMax(int *start,int *end, int **min,int **max) {
    // **min = 3;

You start with initializing your function's argument to point to null so **min is not a defined value ( *min is null but dereferencing a null pointer is UB.) To make **min a valid object, first set *min to point to a valid object, for instance

int *min = malloc(sizeof(int));

(don't forget to deallocate it later) or

int m = 0;
int *min = &m;

or even

int *min = ar;

(thought that latter is already performed in MinMax .)