Question

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Is array name a pointer in C?

So, I usually declare arrays using pointers.

However, you can also declare arrays using square brackets notation:

char a[] = "ok" ;
char b[] = "to" ;
char *pa = a ;

cout << "a " << sizeof( a ) << endl ;   // 3
cout << "pa " << sizeof( pa ) << endl ; // 4

The peculiar thing is, sizeof( a ) will be the actual size of the array in bytes, and not the size of a pointer.

I find this odd, because where is the pointer then? Is a square bracket-declared array actually a kind of datastructure with (sizeof(char)*numElements) bytes?

Also, you cannot re-assign a to b:

a = b ; // ILLEGAL.

Why is that? It seems as though a is the array and not a pointer to the array (“left operand must be l-value” is the error for a = b above). Is that right?

Answer 1

Why is that? It seems as though a is the array and not a pointer to the array (“left operand must be l-value” is the error for a = b above).

a is indeed an array type and not a pointer type.
You cannot assign to an array because it is a non-modifiable lvalue.

BTW Array decays to pointer to the first element when it is passed to a function.