setw(x) & endl;

[Brian_Jones]

#include <iostream>
#include <iomanip>
using namespace std;


int i, j; 


int main()

{
cout << "12345"<< endl;
cout << setw(7)<< endl;
cout << "12345"<< endl;
cout << "12345"<< endl;
cout << "12345"<< endl;
cout << setw(7)<< "K" << endl;
cout << "12345"<< endl;
cout << "12345"<< endl;
cout << "12345"<< endl;

return 0;

}

compiles this:

12345

  12345    
12345
12345
      K
12345
12345
12345
Press any key to continue . . .

In the second box of code, the second set of numbers is indented 2 spaces which is equal to the setw(7). What I don't understand is why it does this. In the first box of code, just having cout << setw(7)<<endl; looks like it does a whole line skip and then continues to use the setw(7) on the next line after that endl; command. I would think the endl; would end any setw but it doesn't appear to. Is there a rule of why it does this?

[tabstop]

I believe the answer to this is that endl is a manipulator and not an object that is actually printed. Since you can "stack" manipulators, the setw doesn't take effect until the next actual print.