On 2007-09-20 05:13, subramanian100in@yahoo.com, India wrote:
> On Sep 19, 2:40 pm, Erik Wikström <Erik-wikst...@telia.com> wrote:
>> On 2007-09-19 08:37, subramanian10...@yahoo.com, India wrote:
>>
>>
>>
>> > Consider the code:
>>
>> > #include <iostream>
>>
>> > using namespace std;
>>
>> > int main( )
>> > {
>> > cout << "test string ";
>> > cout.operator<<(10).operator<<(endl);
>>
>> > return 0;
>> > }
>>
>> > This prints
>> > test string 10
>> > followed by a newline as expected.
>>
>> > However, in main( ), if I have
>>
>> > cout.operator<<("test string
>> > ").operator<<(endl).operator<<(10).operator<<(endl);
>>
>> > it prints
>> > 0x804897c
>> > 10
>> > followed by a newline. It does not print test string but instead
>> > prints its address.
>>
>> > I do not understand.
>> > Does it mean that operator<<(const char *) is not a member function of
>> > ostream ? If so why ?
>>
>> You are correct, but as to why it is not a member I do not know. Last
>> time I saw this discussed none seemed to be able to come up with a good
>> reason, you could try asking in comp.std.c++ though.
>>
>> --
>> Erik Wikström
>
> I just now found a member function
> operator<<(const void * val)
>
> Could the reason be the following :
> If we wanted to print some address, we have to call
> cout.operator<<(ptr) EXPLICITLY for any pointer ptr. So, by calling
> cout.operator<<("test string"), the compiler assumes that we want to
> print the address and so it prints 0x804897c
> When we give cout << "test string", the overloaded non-member function
> is called to print the string literal itself.
>
> Is this reasoning correct ? Excuse me if I am wrong.

Not quite, the only overload of the << operator which takes a pointer is
the one for the char* (or perhaps is was const char*), if you want to
print any other pointer you can use the operator as normal:

#include <iostream>

int main()
{
int* i = new int(1);
std::cout << i;
}

I would assume that the reason for overloading for char* is to allow the
common usage of printing a string literal:

std::cout << "Hello World";

which would not work otherwise.

--
Erik Wikström