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Another simple but confused question...

Posted on 2004-04-16
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Last Modified: 2010-04-01
In the following progrm, I set the precision to 2 initially. That is effective to the whole block.
Then I setn the fill('*'). That is also effective until end of the block after setting once. But why
if that the cout.width(10) is not effective when I set it for one time at the very top? I, infact
have to have the cout.width(10) in every line before cout(ing.)

Tryrunning after commenting out the 2nd, 3rd and 4th cout.width(10), but having just the first one.
Then try uncommenting all cout.width(10).

#include <iostream>
using namespace std;

main()
{
  cout.precision(2);

  cout.width(10);
  cout << 10.12345 << endl;       

  cout.fill('*');
  cout.width(10);
  cout << 10.12345 << endl;       

  //field width applies to string too.
  cout.width(10);
  cout<<"Hi!" << endl;            

  cout.width(10);
  cout.setf(ios::left);       

  cout << 10.12345;            

  return 0;
}
0
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Question by:prain
  • 2
7 Comments
 
LVL 16

Accepted Solution

by:
nonubik earned 25 total points
ID: 10842882
operator << sets the width to 0. (that's its implementation in ostream.h)
0
 
LVL 12

Expert Comment

by:stefan73
ID: 10842968
Hi nonubik,
> operator << sets the width to 0. (that's its implementation in ostream.h)

Yes. Try this:
  cout.width(10);
  cout << 10.12345 <<endl;      
  cout << 10.12345 <<endl;      


Cheers,
Stefan
0
 
LVL 6

Assisted Solution

by:Mafalda
Mafalda earned 25 total points
ID: 10843866
You could use this

#include <iomanip>
using std::setw;
using std::setfill;

cout.precision(10);
cout << setfill('*') << setw(10) << 10.12345 << endl << setw(10) << 10.12345 << endl << setw(10) << 10.12345 << endl;
0
 
LVL 16

Expert Comment

by:nonubik
ID: 11134718
I think I gave a straight ansewr....
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