Absurd Program Output

[C-RaT]

Hello all
i tried the following code and got unexpected outputs.
can anyone provide me with any feedback on any mistake that i have done.

#include<iostream.h>
#include<stdio.h>
void stm()
        {
                cout<<"hello";
        }
struct a
        {
                stm();
        }b;
main()
        {
              cout<<endl<<b.stm();
              cout<<endl<<sizeof(b)<<endl;
              stm();
              return(0);
          }

output :

[

it do not compiles why ?
and on commenting the first cout it gives the value 1.

[tabstop]

No such thing as iostream.h, although there is a header called "iostream". Similarly, no such thing as stdio.h, although there is a header "cstdio" -- but you don't need it for this program, since you're not using any functions from that header.
You're missing a namespace std declaration, to use cout and endl without qualification.
You can't call a function inside a struct, and you can't declare a function without giving a return type, so that stm() inside the struct declaration is just wrong.
And finally, struct a contains no members, so sizeof b must be at least zero, but it can be larger.

[hk_mp5kpdw]

#include<iostream>
#include<cstdio>

void stm()
{
    std::cout << "hello";
}

struct a
{
    void stm();
}b;

int main()
{
    std::cout << std::endl << b.stm();
    std::cout << std::endl << sizeof(b) << std::endl;
    stm();
    return(0);
}

You're trying to cout the value of b.stm() which is a function call returning a void, you can't do that.

You can't call a function inside a struct

Sure you can, ala struct/class member functions. ...not in the way the OP's trying to do it here but it can be done.

And finally, struct a contains no members, so sizeof b must be at least zero, but it can be larger.

I think "empty" structs need to be at least 1 byte.

[Elysia]

Sure you can.

I think tabstop is referring to the fact that you can't call a function inside a struct declaration (or definition). It must be done inside a function inside the struct - ie the implementation, not in the definition.

[C-RaT]

Tabstop:

as far as i know if i dont declear any namespace its the default namespace that is taken.
and please google for "c++ headers" you'll be able to find both of them i used .
and i am using "cout" so i think iostream.h would be used atleast.

[tabstop]

Sure you can... not in the way the OP's trying to do it here but it can be done.

Sorry, should have been clearer -- you can't call a function from the definition of a struct -- you can call a method of a struct, I agree.

[Elysia]

as far as i know if i dont declear any namespace its the default namespace that is taken.

cout is not located in the default namespace, it's located in std. So are any other standard library functions.

and please google for "c++ headers" you'll be able to find both of them i used .

Who do you believe more? Some old web pages or expert programmers?
Those headers are deprecated and do not exist. They are not standard headers.

and i am using "cout" so i think iostream.h would be used atleast.

Make that "iostream".

[anon]

You probably need a more modern compiler. It seems that you are using something that came out before C++ was standardized (end of 1990s).

Here's what your program would look like in C++.

//proper header
#include<iostream>
using namespace std;

void stm()
    {
        cout<<"hello";
    }

struct a
    {
        //functions require explicit return type
        int stm()
        {
            //if you want to keep the same name
            //you have to indicate that you mean the global stm function
            ::stm();
            //if you declare your functions to return something,
            //you actually have to return something
            return 42;
        }
    }b;

//functions require explicit return type
int main()
    {
      cout<<endl<<b.stm();
      cout<<endl<<sizeof(b)<<endl;
      stm();
      return(0);
    }