what does cout return?

what does cout return?In code

Code:

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cout<<cout<<"Hello World";

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I am getting a hexadecimal number,probably an address.Does cout return something?Thanking in advance

cout doesn't return anything. The '<<' operator returns cout.

When you write (cout << "x" << 3 << " crazy!" << endl), evaluation happens like this:

Code:

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((cout << "x") << 3) << " crazy!") << endl
((cout << 3) << " crazy!") << endl
(cout << " crazy!") << endl
cout << endl
cout

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As written, I'd say you're outputting the address of a function.

Which function's address?.I havenot made any function yet.

Code:

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#include<iostream.h>
int main()
{
cout<<cout<<"Hello world";
return(0);
}

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In above code i have not made or called any function.So which function's address are you taking of.

>I havenot made any function yet.
main is a function, so this statement is false.

>what does cout return?
cout is an object, it doesn't return anything. The << operator returns a reference to the stream to which it's applied, in this case cout. So the address you're seeing is that of cout.

Quote:

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cout is an object, it doesn't return anything. The << operator returns a reference to the stream to which it's applied, in this case cout. So the address you're seeing is that of cout.

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I thought that also but its not :)
Consider:

Code:

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#include <iostream>

using namespace std;

int main()
{
        cout<<cout<<"Hello World"<<endl;
        void* ptr = cout;
        cout<<ptr<<endl;
        cout<<(&cout)<<endl;

}

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The output recieved (obviously the exact addresses will be different by machine)

Quote:

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0x804de9cHello World
0x804de9c
0x804de98

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So it appears that cout<<cout; returns a void pointer address.
Using gdb I was able to confirm that it does in fact call
std::basic_ios<char, std::char_traits<char> >::operator void*() const

Quote:

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or called any function.

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the << operator calls operator<< for whatever datatype you pass it so you did call a function.

Take for example overloading the << operator . Might make it seem a little clearer.

Code:

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Class complex{};       
std::ostream& operator<< (std::ostream&, const Complex&);

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this is in main of course

Code:

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Complex b;
cout << b;
//which is equivalent to the call
cout.operator(b)

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So in fact you are calling a function, in this case the overloaded << function

Quote:

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Originally Posted by vaibhav

what does cout return?In code

Code:

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cout<<cout<<"Hello World";

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I am getting a hexadecimal number,probably an address.Does cout return something?Thanking in advance

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You might try reading the manual (scroll down the page to the "return value" section). The return value of a stream is well documented in every c++ textbook and millions of online references. If you don't know what something does, then look it up in the manual before you hurt yourself. :rolleyes:

Quote:

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Originally Posted by Ancient Dragon

You might try reading the manual (scroll down the page to the "return value" section). The return value of a stream is well documented in every c++ textbook and millions of online references. If you don't know what something does, then look it up in the manual before you hurt yourself. :rolleyes:

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You might want to try reading the thread. The question wasn't what operator<< returned but what cout<<cout displayed. Yea they used return wrongly but thats pretty obvious if you read the post and the replies. Oh and next time you link something at least make sure its relevent. (you linked operator>> which is istreams and we are talking about operator<< which is ostream)

Code:

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#include <iostream>

using namespace std;

int main()
{
        cout<<cout<<"Hello World"<<endl;
        void* ptr = cout;
        cout<<ptr<<endl;
        cout<<(&cout)<<endl;

}

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I compiled above program.I get output as

Code:

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0x8f910378Hello World
0x8f910378
0x8f910374

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There is one similarity in yours and my output. address returned by cout is exactly 4 bytes greater than address of cout.I think it may help you in finding what cout returns?

My teacher replied this as

Code:

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In general, s<<a where s is a stream returns a reference to s.

cout<<cout; converts the second "cout" to a printable value using the
conversion
void* that we use for testing its state (see your C++ textbook). That
value is
only guaranteed to be zero (meaning the state is bad) or non-zero
(meaning the
state is good). Nothing more is guaranteed.

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Can anyone explain this to me in simple language?

See this

Quote:

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Originally Posted by Ancient Dragon

See this

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Borland bcc32 gives the following output (different in nature from the others, but compliant with the actual spec):

Quote:

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00000001Hello World
00000001
0042049C

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D

Thanks all of you ,Now I understands