请解释输出？

Question

can anyone explain the strage output of the program I know that the value has nothing to do with the value stored in the array but with the pointer thing but how is the second value coming to be 5:

int main()
{
    int **h;
    int a[2][2]={1,2,3,4};
    h=(int **)a;
    int i,j;
    printf("%d",*h);
    (*h)++;
    printf("\n%d",*h);

    getch();
    return 0;
}

Answer 1

What's happening is that *h is of type int* which is a pointer.

When you increment it will actually increment by 4 rather than 1. Therefore the number you print out in the end is 1 + 4 = 5 .

Here's your code with more prints:

  int **h; 
  int a[2][2]={1,2,3,4}; 
  h=(int **)a; 

  cout << h[0] << endl;
  cout << h[1] << endl;
  cout << h[2] << endl;
  cout << h[3] << endl;

  int i,j; 
  printf("%d",*h); 
  (*h)++; 
  printf("\n%d",*h); 

  cout << endl;
  cout << h[0] << endl;
  cout << h[1] << endl;
  cout << h[2] << endl;
  cout << h[3] << endl;

The output is:

00000001
00000002
00000003
00000004
1
5
00000005
00000002
00000003
00000004

So you can see the first value, being incremented by 4. Because 4 is the size of the pointer when compiled for 32-bit.

Answer 2

By the statement h=(int **)a; you have only allocated the memory address(the first) of the array a to h . You have also defined h as a pointer to a pointer which can also point to a two-dimensional array as you have done. Also, to be seen, you have not made h a two-dimensional array (using malloc recursively). By printf("%d",*h); , you are trying to access the value at the address stored in h .

Arrays store values in memory in two ways, either column-wise or row-wise . In either ways the memory locations are sequential, in your case also. So, h stores the memory address of the first element of the array a . Therefore, when you use *h it retrieves the value at the address stored in h , ie, the first value in array a . And when you increment *h by (*h)++ , it increments because *h is still a pointer, and as you know incrementing a pointer doesn't mean incrementing by 1 , it actually increments by 4 .

Hence, you are getting the above-mentioned output of yours.

Further discussions are welcomed from you Ankit .

- Sandip

Answer 3

Since the expression *h is a pointer type, pointer arithmetic comes into play. Remember that pointer arithmetic takes the size of the base type into account; for any pointer of type T *p , the expression p++ will advance the pointer p by sizeof T bytes.

Since *h is a pointer initialized to the value 1, the expression (*h)++ is read as "add sizeof (int *) bytes to 1", which in your case is obviously 4. Hence the output of 5.