2Dimensional Array Pointer manipulation in C++

Question

int main(){
    int a[10][10];
    int **ptr =(int **)a;
    cout<<a<<endl<<ptr<<endl;
    cout<<*a<<endl<<*ptr<<endl;
    return 0;
}

Output of this code on my computer is

0021FC20
0021FC20
0021FC20
CCCCCCCC

Why is "a" equal to "*a"? why isn't *a equal to *ptr?

Answer 1

Why is a equal to *a?

When used in a context that requires a pointer, an array will be converted to a pointer to its first element. a is an array of arrays; so it will decay to a pointer to the first array. *a is the first array, and will decay to a pointer to the first integer in that array. Both of these exist at the same location, so the two pointers will have equal values.

why isn't *a equal to *ptr?

Because the conversion from an array of arrays, to a pointer-to-pointer, is not valid. You have forced the conversion using a cast - the dangerous C-style cast, which in this case acts like reinterpret_cast - so *ptr will read the first few bytes of the array and interpret that as a pointer (probably - the behaviour here is undefined, so in principle anything could happen). There is no pointer in the memory pointed to by ptr, so *ptr will certainly not give you a valid pointer.

Answer 2

Why is a equal to *a?

Since you cannot print an array, a is implicitly converted from int[10][10] to int(*)[10]. So what actually gets printed instead of a is a pointer to the first line of a.

*a is the first line of the array, and that in turn gets converted to a pointer to the first element.

Since an array has the same address as its first element, you get the same value twice.