From: Daniel Finnie <danfinnie@...>
Date: 2007-03-20T10:27:53+09:00
Subject: Re: C(++) For Loop Equivalents

Use the step method:

----------------------------------------------------------- Numeric#step
      num.step(limit, step ) {|i| block }     => num
------------------------------------------------------------------------
      Invokes _block_ with the sequence of numbers starting at _num_,
      incremented by _step_ on each call. The loop finishes when the
      value to be passed to the block is greater than _limit_ (if _step_
      is positive) or less than _limit_ (if _step_ is negative). If all
      the arguments are integers, the loop operates using an integer
      counter. If any of the arguments are floating point numbers, all
      are converted to floats, and the loop is executed _floor(n +
      n*epsilon)+ 1_ times, where _n = (limit - num)/step_. Otherwise,
      the loop starts at _num_, uses either the +<+ or +>+ operator to
      compare the counter against _limit_, and increments itself using
      the +++ operator.

         1.step(10, 2) { |i| print i, " " }
         Math::E.step(Math::PI, 0.2) { |f| print f, " " }

      _produces:_

         1 3 5 7 9
         2.71828182845905 2.91828182845905 3.11828182845905

So your example would become:
0.step(i*j/2, 1) # (I'm not too sure about the details of C++ and if 
defining i in the beginning would make i*j/2 be 0 no matter what)
Or, you could do
0.upto(i*j/2) # assumes step is one.

Finally, you could do
for i in 0..(i*j/2) do |i|
	# blah...
end

Daniel Sheppard wrote:
>> What is the most efficient way in Ruby to create a conditional loop in
>> Ruby that looks like this in C(++):
>>
>> for(int i=0; i*j/2<maxNum; i++)
>> {
>>   //cool method stuff
>> }
> 
> In the general case, you can 'while' anywhere where you'd normally 'for'
> 
> i=0
> while i*j/2<maxNum 
>   //cool method stuff
>   i++
> end
> 
> But 'most efficient way' to do a conditional loop depends on your
> context (and also what you're trying to make efficient - runtime?
> writetime? readtime?).
> 
> Dan.
> 
>