C & C++: What is the difference between pointer-to and address-of array?

Question

C++11 code:

int a[3];
auto b = a;       // b is of type int*
auto c = &a;      // c is of type int(*)[1]

C code:

int a[         


        

Answer 1

Consider this example:

#include

int main()
{
    int myArray[10][10][10][10]; //A 4 Dimentional array;

    //THESE WILL ALL PRINT THE SAME VALUE
    printf("%d, %d, %d, %d, %d\n",
            myArray,
            myArray[0],
            myArray[0][0],
            myArray[0][0][0],
            &myArray[0][0][0][0]
          );

    //NOW SEE WHAT VALUES YOU GET AFTER ADDING 1 TO EACH OF THESE POINTERS
    printf("%d, %d, %d, %d, %d\n",
            myArray+1,
            myArray[0]+1,
            myArray[0][0]+1,
            myArray[0][0][0]+1,
            &myArray[0][0][0][0]+1
          );
}

You will find that all the 5 values printed in first case are all equal. Because they point to the same initial location.

But just when you increment them by 1 you see that different pointers now jump (point) to different locations. This is because myArray[0][0][0] + 1 will jump by 10 integer values that is 40 bytes, while myArray[0][0] + 1 will jump by 100 integer values i.e by 400 bytes. Similarly myArray[0] + 1 jumps by 1000 integer values or 4000 bytes.

So the values depend on what level of pointer you are referring to.

But now, if I use pointers to refer all of them:

#include

int main()
{
    int myArray[10][10][10][10]; //A 4 Dimentional array;

            int * ptr1 = myArray[10][10][10];
            int ** ptr2 = myArray[10][10];
            int *** ptr3 = myArray[10];
            int **** ptr4 = myArray;

    //THESE WILL ALL PRINT THE SAME VALUE
    printf("%u, %u, %u, %u\n", ptr1, ptr2, ptr3, ptr4);

    //THESE ALSO PRINT SAME VALUES!!
    printf("%d, %d, %d, %d\n",ptr1+1,ptr2+1,ptr3+1,ptr4+1);
}

So you see, different levels of pointer variables do not behave the way the array variable does.