Programming

Question

int main()
{
    int array[5][5];
    printf("%d",( (array == *array) && (*array == array[0]) ));
    return 0;    
}

Here I understand *array=*(array+0)=array[0]

but how is array=*array?

*array contains the address of the 0th element of 0th 1-D array whereas array contains the address of 0th 1-D array , which are not same according to my knowledge. Then why is the output 1 here?

Can someone pls help me out here..

Comments

unlike pointers, array name is just mnemonic and therefore does not occupy memory,

therefore 'array' will have same address as '*array' which in turn will have same address as '**array'

---

Considering 1-D array,

the array name is a constant pointer which stores the base address of the array, right? and when we do *array we get the value of the 1st element..

So here array gives us an address whereas *array is a value at that address?

---

yes, indeed.

---

array[0] = *(array+0) is correct.

you are doing wrong.

---

@nitish

In a 3-D array when I execute these statements they all give same output. Can u pls brief me why, it will help clear my doubts..

printf("%p\n",arr);
printf("%p\n",arr[0]);
printf("%p\n",arr[0][0]);