why it shows me "1"

[phongvants123]

it's always show me "1"....can you help me

#include<iostream>
using namespace std;
int b();
int main(){
cout<<b;
return 0;
}
int b()
{
cout<<"need to show on consolve";
}

[whitenite1]

Change line 5 to cout << b();
You are not calling the function, doing it your way, but are displaying an non-initialized variable.

[phongvants123]

you are definately right...can you help me one more time ?

[andywestken]

Actually, what cout << b; does is write out the address of the function b()

[phongvants123]

i know i'm wrong when i type

cout<<b;

but can you explain why does it show me

1

anytime

[mbozzi]

Actually, what cout << b; does is write out the address of the function b()

No: the name b is ultimately converted to bool and printed. That is, the overload std::ostream::operator<<(std::ostream&, void*) isn't applicable and
std::ostream::operator<<(std::ostream&, bool) is called instead.

This is because function pointers aren't convertible to void*.

Since the address of a function will never be NULL (i.e., 0), it is treated as true and printed as 1.

[phongvants123]

it is a little bit higher than i thought....can not get it after all....anyway i appreciate that

[Ganado]

phongvants123, I'm not sure what your last post meant, but your b() function is supposed to return an int. Right now you don't have it returning anything.

If you have no use for an int, maybe return a string?

You could change your function to something like

#include <string>
#include<iostream>

using namespace std;

string b();

int main()
{
    cout << b() << endl;
    return 0;
}

string b()
{
    return "need to show on console";
}

[phongvants123]

i mean what is the diference between

cout<<b;

and

cout<<b();

[jonnin]

b is a variable. (in this case the it is the name of a function, which is 'sort of' a variable in some sense of the word, at the compiler/ assembler level).
b() is a function call result.

printing b (which is a function) here is going to give the address of the function, I think.
b() is correct.