Why is it that, when I dereference an array pointer, the resulting value is a pointer to the first element of the array, not the entire array object?

Question

#include<iostream>

int num[3]={66,77,88};

int main()
{
    int(*pi)[3]=&num;
    std::cout<<*pi;
} 

The result is an address instead of an array. What is the explanation behind this?

Answer 1

not the entire array object?

*pi gives the array, i.e. the int[3]. But operator<< for std::basic_ostream doesn't have overloads taking array but has an overload taking pointer (const void*), then array-to-pointer decay happens, the converted pointer (int*) is pointing to the 1st element of the array and then is converted to const void*, then is passed to std::cout and gets printed out.

There is an implicit conversion from lvalues and rvalues of array type to rvalues of pointer type: it constructs a pointer to the first element of an array. This conversion is used whenever arrays appear in context where arrays are not expected, but pointers are:

Answer 2

Just like cout cannot print a vector, it cannot print an array. However cout will print the address of the first element.

If you take a look at the source code of iostream they mention cout as a pointer https://code.woboq.org/llvm/libcxx/src/iostream.cpp.html, and that is also why you always need to iterate over each item of your array as oppose to other languages (like Rust) where you can print a vector or an array.

For string there is an exception since in C and C++ they are (or should be!) NULL terminated, and that is why the whole char array is printed. But 0 is a valid number, hence there is no way to know the bounds of an int array unless you can detect it at compile time (which Rust does).