How does the setw stream manipulator work?

Question

How does the setw stream manipulator (space count) work? When there is one \t, for example, I want to print a with four spaces, so I use \t and I compare \t with setw.

The code that I wrote:

# include <iostream>
# include <iomanip>

int main()
{
    std::cout<<"\t"<<"a\n";
    std::cout<<std::setw(9)<<"a\n";
    return 0;
}

Output:

   a // This is 1 '\t'
   a // This is setw()

So what I thought was:

setw(18) = \t\t

The code worked. But when I deleted the \n, it did not become one straight line.

# include <iostream>
# include <iomanip>

int main()
{
    std::cout<<"\t\t"<<"a\n";
    std::cout<<std::setw(18)<<"a";
    return 0;
}

It gives me this output:

      a
       a

What's wrong?

Answer 1

That's because you need to add a \n at the setw(18). And this applies to any setw.

Sample code:

# include <iostream>
# include <iomanip>
int main()
{
std::cout<<"\t\t"<<"a\n";
std::cout<<std::setw(18)<<"a\n"; // And you add the \n here
return 0;
}

Output:

        a
        a

And another solution is:

# include <iostream>
# include <iomanip>
int main()
{
std::cout<<"\t\t"<<"a\n";
std::cout<<std::setw(18)<<"a "; // And you add the a space here
return 0;
}

And the output will be the same.

---

The reason behind why should we put the \n or a space is because:

It's because it justifies the whole two-character string "a"\n, not only the a. If you print the newline separately (... << 'a' << '\n') then you would get the same "error". You could also have solved it by using almost any space character, like ... << "a "; You might want to print both with and without the newline in the same program to see the difference. - Some Programmer Dude