quick c++ question

Factual Questions
1

basically i’m trying to make 2 variables output to each other left justified with only a space between them…

cout << setw(20) << fname1 lname1

I don’t know what goes between fname1 and lname1 to make that work. But i basically want lname1 to left justify to fname1 with a space between and be within the confines of the 20 space width.

Can anyone help me out?

2

What if lname1, fname1, or the concatenation of the two is longer than 20 characters?

cout << setw(20) << fname1 << ’ ’ << lname1 << flush; should do something reasonably close to what you want to do, I think. I usually use stdio instead of streams, so I’m not 100% sure.

3

C++ streams are somewhat unwieldy for formatted output; I prefer the C *printf functions. If you insist on C++ streams, though, here’s one way:


  ostringstream os;
  os << fname1 << " " << lname1;

  cout << setw(20) << left << os.str();

This will left-justify the concatenation of fname1 and lname1 (with one space separating the two) in a 20-character field, expanding if necessary. (The setw() affects only the next value to produce output. The left sets left justification (standard is right). For these you may need to #include .)

A C solution, which I think is legal :slight_smile: :


  printf( "%-*s", 20-printf( "%s ", fname1 ), lname1 );

ultrafilter’s solution will set fname1 right-justified in a 20-character field, with lname1 following it and not justified.