CODEWITHSUNDEEP
c++
S.A.Parkhid
S.A.Parkhid

Reputation: 2870

What is the difference between int and char arrays?

What is the difference between int and char arrays below: 

int main()
{
    int numbers[] = {2,1,3};
    char letter[] = {'a','b','\0'};
    cout<< numbers<<endl;
    cout<< letter<<endl;
}

Output: 

0x22ff12 // an address
ab

Why isn't the 213 displayed ? I know the name of an array will point to the address of its first element, but why does a char array display different behavior?

Upvotes: 6

Views: 4397

Answers (10)

Nikhil Singh
Nikhil Singh

Reputation: 1

A char array contains characters.
It can be initialized like:

char arr[4]={'a','b','c','\0'};
char arr[4]={"abc"};

An integer array contains integers.
It can be initialized like:

int arr[4]={1,2,3,4};

Upvotes: 0

Pete Kirkham
Pete Kirkham

Reputation: 49311

In C, and therefore in C++, a string is often represented by an array of chars terminated in a 0. Therefore an overloaded operator<< is provide for the class std::ostream, of which std::cout is an instance , which prints the char* as a string. There is no such common use of int arrays, nor any convention so the operator would 'know' how many elements to output, so the pointer to the array is matched to the version of operator<< which outputs any other pointer by printing its address.

Upvotes: 4

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361422

The reason is thatoperator<< overloaded for const char* which prints each character till it encounters \0.

There is no such overload corresponds to int[N] which prints each element in it. Instead when you write cout << numbers, it invokes operator<< which is overloaded for void*, and which prints the address.

However, if you overload operator<< for T[N], then you can print it like that as well.

Here is a simple illustration:

template<typename T, size_t N>
std::ostream & operator<<(std::ostream & out, const T (&a)[N])
{
   for(size_t i = 0 ; i < N ; ++i)
      out << a[i] << ' ';
   return out;
}

int main()
{
   int numbers[] = {2,1,3};
   char letter[] = {'a','b','\0'};
   cout<< numbers<<endl;
   cout<< letter<<endl;
}

Output:

2 1 3 
a b

Demo : http://ideone.com/O4T9N

Upvotes: 6

John Dibling
John Dibling

Reputation: 101456

There is no operator<< overload that takes arrays, exactly, so the arguments you pass (eg numbers and letter) undergo array-to-pointer conversion, to void* and char* respectively.

There is an overload of operator<<() that takes a const void*, and another that takes a const char*. When you call:

cout<< numbers<<endl;

the const void* version is matched, but when you call:

cout<< letter<<endl;

the const char* version is matched.

In the const void* version, the pointer is displayed, while with the const char* version, the string is displayed up to the null terminator.

Upvotes: 11

Jeff Burdges
Jeff Burdges

Reputation: 4261

There is no reason the compiler should know where your int[] array ends, but tradition and standard libraries dictate that C strings are null terminated char[] arrays. There is no such tradition or library support for null terminated int[] arrays.

There are C++ pretty printers templates available if you need this functionality. I vaguely recall that one employs an array's bound when the type actually knows the bound, i.e. your code still won't work since you use [] not [3].

Just fyi, your code cannot be fixed by replacing the [] with a [3] inside the STL, although perhaps operator<< could be overloaded.

Upvotes: 0

Zaur Nasibov
Zaur Nasibov

Reputation: 22659

Your code mostly refers C. An array of char is de-facto representation of strings in C. On the other hand, an array in C is also a pointer to the memory cell (an address of the cell) that holds the first element of the array.

So, when you print out an array of characters, you in fact print out a string (because C treats it that way). When you're printing an array of integers, you're printing out the address of the first element of the array.

Upvotes: 1

Emil Lundberg
Emil Lundberg

Reputation: 7380

numbers is a pointer. All arrays in C++ are in fact pointers, numbers[3] just means "the value at the memory address &number+3", so you're outputting the memory address of the first element in numbers.

Upvotes: 0

Bo Persson
Bo Persson

Reputation: 92261

char arrays are special because there is an overload for operator << that displays the content as a string.

All other arrays will have the address displayed by default.

Upvotes: 3

Mysticial
Mysticial

Reputation: 471229

When you print an array with cout it will print the base address of the array.

The exception is with char arrays which have been overloaded to print it as a c-string.

If you want to print the elements of the int array, you need to do it element-by-element.

Upvotes: 9

Tudor
Tudor

Reputation: 62439

In C/C++ an array is in fact a pointer to the first element. A pointer holds the address where a value is stored. Therefore, if you print the pointer numbers, you will get the address where the first value (2) is stored in memory.

char* is an exception, as it will behave as a string when you try to print it.

Upvotes: 1

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