From: Brian Candler <B.Candler@...>
Date: 2007-02-12T20:53:52+09:00
Subject: Re: embedding if statements in upto function.(newbie question)

On Mon, Feb 12, 2007 at 08:05:01PM +0900, Michael Sc wrote:
> Thank you to everyone for all of your help.
> I was trying to use upto to go through an array of arrays while using if
> statments.  Could anyone enlighten me on what I am doing wrong with the
> if statements?
> I am sure it is something easy.
> 
> I tried this:
> 0.upto( a.size-2 ) {|i|
>   if (a[i][2]==a[i+1][2] and a[i][0]==a[i+1][0]) then
>     print { (a[i+1][3] - a[i][3]) / a[i][3] * 100 } /n
> else print 0
> end}

Why don't you give some sample data, and show what happens when you run it.
Preferably make a small, standalone program which demonstrates the problem.
Do you get an exception raised? A wrong result?

Looking at the above, the fact that you have used curly brackets in the
'print' statement certainly won't help... this creates a code block and will
give you a confusing error like this:

irb(main):003:0> print { 4 + 5 } / 3
nilNoMethodError: undefined method `/' for nil:NilClass
        from (irb):3
        from :0

This is equivalent to:

print( { 4+5 } ) / 3
      ^   ^     ^
      |   |     |
      |   |   value returned by
      |   |   print() is nil
      |   |
      | a code block
      | which is not used
      |
    no arguments
    to print

BTW you probably want to use puts rather than print, so your numbers are on
separate lines; or at least print a space between them.

Your above code does also highlight a skeleton in Ruby's closet: note that

    puts (4+5)/3

    puts(4+5)/3

are two different constructs, even though they differ only by one space.
They are equivalent to, respectively,

    puts( (4+5)/3 )           # probably what you want

    ( puts(4+5) ) / 3         # prints 9, tries to divide nil by 3

B.