About cout

[tuyongce]

Source code:
#include <iostream>
using namespace std;

int main()
{
cout.operator<<("test string");

return 0;
}

The output as follow:
0046B01C ----> why it's an address?

My question:
1) Why the output was an address and why not "test string" as expected
?
2) What's the difference between
cout<<"test";
and
cout.operator<<("test");
?
(I had thought they were the same! )

 

[benben]

Source code:
#include <iostream>
using namespace std;

int main()
{
cout.operator<<("test string");

return 0;
}

The output as follow:
0046B01C ----> why it's an address?

My question:
1) Why the output was an address and why not "test string" as expected
?
2) What's the difference between
cout<<"test";

The equivalent of the above should be:

std:perator<< (std::cout, "test");

and
cout.operator<<("test");

This only dumps an address.

 

[eriwik]

Source code:
#include <iostream>
using namespace std;

int main()
{
cout.operator<<("test string");

return 0;

}The output as follow:
0046B01C ----> why it's an address?

My question:
1) Why the output was an address and why not "test string" as expected
?
2) What's the difference between
cout<<"test";

This calls this function:
basic_ostream<charT, traits>&
operator<<(basic_ostream<charT, traits>& os , const

and
cout.operator<<("test");

This (I think) calls this function:
basic_ostream<charT,traits>&
operator<<(basic_streambuf<char_type,traits>* sb );

 

[Ron Natalie]

My question:
1) Why the output was an address and why not "test string" as expected
?
2) What's the difference between
cout<<"test";
and
cout.operator<<("test");
?
(I had thought they were the same! )

operator<< doesn't necessarily have to be implemented as a member
function.

By using the explicit call to the member function, you won't
match any of the nonmember function overloads.

 

[Rolf Magnus]

Source code:
#include <iostream>
using namespace std;

int main()
{
cout.operator<<("test string");

return 0;
}

The output as follow:
0046B01C ----> why it's an address?

My question:
1) Why the output was an address and why not "test string" as expected
?

There is no perfectly fitting operator in the class, so the type is using
the conversion that fits best.

2) What's the difference between
cout<<"test";

Here, the compiler can chosse between member an non-member operators.

and
cout.operator<<("test");
?

Here, the operator must be a member. Only few of the streaming operators for
std::cout are members.