cout a stringstream but print a pointer

Question

This is my code:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

class TestLog: public std::stringstream
{
        public:
                ~TestLog()
                {
                        cout << (str()) << endl; // Why does this print an address ?
                }
};

int main()
{
        TestLog() << "Hello World!"; //test 1 print an address
        stringstream ss;
        ss << "Hello World!";
        cout << (ss.str()) << endl; //test 2 print a string

        return 0;
}

And the output:

0x401b90

Hello World!

Compiler info :

g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-11)

In my opinion, (a)the str() method of std::stringstream return a string. (b)std::cout is an object of std::ostream. So both of the two test will call the same operator funtion of ostream and print same "Hello world". But test 1 print an address , test 2 print the correct "Hello world".
What wrong with me ? Thanks.

[can't reproduce](http://melpon.org/wandbox/permlink/x6C0v3XtW00OToKY) — W.F., Jan 02 '17 at 10:50
[Ideone](http://ideone.com/6kdI2E) prints correct results too. — Quest, Jan 02 '17 at 10:52
[There we go](http://melpon.org/wandbox/permlink/QwSPL1PdLvm2XW0r). The behaviour arises when compiling as C++03. I added the tag for you. — Quentin, Jan 02 '17 at 10:55
Thank you. I test it with Vs ,and both output are "Hello world". — 杜智超, Jan 02 '17 at 10:57

Quentin · Answer 1 · 2017-01-02T13:07:31.690

ss << "Hello World!"; is resolved to a call to the following overload (#2 on this page):

template <class Traits>
std::basic_ostream<char, Traits> &std::operator << (
    std::basic_ostream<char, Traits> &os, char const *s
);

This overload decays the string literal into a char const *, then prints it.

To stir the mud, we can try the following snippet:

TestLog tl;
tl << "Hello World!";

This one will also resolve to the overload above, and print Hello World!. That is because tl is an lvalue which can bind to the first parameter that is a non-const lvalue reference.

In your example, TestLog() is an rvalue -- this overload cannot match! Thus another overload is chosen instead (#7 here):

std::basic_ostream &std::basic_ostream::operator << (void const *value);

This one is a member function overload, and has been inherited from std::stringstream. Even though you can't bind a non-const reference to an rvalue, you can call non-const member functions on an rvalue. So this overload is a valid match, and it is chosen -- printing the address of the literal as if it were any old pointer.

C++11 brought a new overload to solve this issue, visible at #3 here:

template <class CharT, class Traits, class T>
std::basic_ostream<CharT, Traits> &operator << (
    basic_ostream<CharT, Traits> &&os, T const &value
);

T const & matches the literal's char const[N] type perfectly, so it is ranked higher than the void const * overload. The rvalue reference as first parameters binds to the temporary just fine.

Named rvalue references are considered lvalues, so this function can then call os << value; again to trampoline back into the overload set for an lvalue stream. Thus, in C++11 and above, both lines print Hello World!.

cout a stringstream but print a pointer

1 Answers1