C ++：数组和内存位置

Question

Array of characters has always managed to intrigue me.

Lets create an array of integer type-

int a[]={1,2,34};
cout<<a;

in this case cout<<a gives us the location of the array a

But-

char a[]="C++";
cout<<a;

gives us C++ as the display , it is quite interesting.

1) So why doesnt the array of characters show its memory location when used with cout?

2) How can I find the location of an array of characters?

All help would be appreciated :)

- Newbie in C++ with a will to learn

Answer 1

"1) So why doesnt the array of characters show its memory location when used with cout?"

There's a specialized overload for

std::ostream& operator<<(std::ostream&, const char*);

that's why you always see the characters printed, not the address.

"2) How can I find the location of an array of characters?"

You can cast to a void* pointer if you want to see the address of the char array:

char a[]="C++";
cout << (void*)a;

or just prefix the address operator as @WhozCraig suggested in their comment

cout << &a;

Answer 2

Below are two overloads of operator<< available in standard:-

std::ostream& operator<<(std::ostream&, const char*);  // _1
std::ostream& operator<<(std::ostream&, void*);  //       _2

int arr[] = {1,2,3};
cout << arr;             // This would invoke _2

char arr[] = "Hello";
cout << arr;             // This would invoke _1

char arr[] = "Hello";
cout << (void*)arr;      // This would invoke _2
cout << &arr;

Hope I'm clear

Answer 3

First of all rake into account that there is not operator << for integer arrays.

When you write statements like these

int a[] = { 1, 2, 34 };
cout << a;

then expression a is implicitly converted to a pointer of type const void * that to select the following overloaded operator

basic_ostream<charT,traits>& operator<<(const void* p);

As for character arrays then the C++ Standard explicitly defines overloaded operator << for type const char *

template<class traits>
basic_ostream<char,traits>& operator<<(basic_ostream<char,traits>&,
const char*);

If you want that for character arrays would be called the operator << that is defined for type const void * then you have to explicitly cast the pointer to thjis type. For example

char a[] = "C++";
cout << static_cast<const void *>( a );