why is macro acting different while not using namespace std and while using namespace std?

Question

in this below code macro 1 is always fine

but, macro 2 is not working if statement 1 is not written..why is this happening?

#include<iostream>
#include<conio.h>
//using namespace std;      //--statement 1
#define l std::cout<<       //--macro 1
#define nl std::cout<<endl; //--macro 2
int main(){
      l "testing";  
      nl  // this is not working if i dont use statement 1 
      l "a new line";
getch();
return 0;
}

when statement 1 is not written macro 2 is producing a error stating that '[Error]endl was not declared in this scope'

if cout<< is the short version of std::cout<<, this error should not happen...i can not understand why is this happening...

you are just missing the std:: before endl. It should be **std::endl;** — banetl, Nov 17 '16 at 09:21
`cout` is not a "short version" of `std::cout`. It *is* `std::cout`, once you've dumped it into the global namespace via `using namespace std;`. — Quentin, Nov 17 '16 at 09:22
[don't use `endl` unless you really need its side effect](http://stackoverflow.com/q/213907/995714) — phuclv, Nov 17 '16 at 09:23

score 0 · Accepted Answer · answered Nov 17 '16 at 09:26

0

This is not the question of your MACRO.

The endl object also belongs to the std namespace, therefore either you should use

std::cout << std::endl;

or

using namespace std;

cout << endl;

answered Nov 17 '16 at 09:26

nyekiz

91
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why is macro acting different while not using namespace std and while using namespace std?

1 Answers1