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comp.lang.ruby

Math Tricks

James Gray

9/29/2004 10:09:00 PM

Quick question. When I run:

#!/usr/bin/env ruby

class RPNCalc
def initialize(stack = [ ])
@stack = stack
end

def push(number)
if number.kind_of? Numeric
@stack.push(number)
return @stack[-1]
else
return nil
end
end

def add; return binary { |l, r| l + r } end
def sub; return binary { |l, r| l - r } end
def mul; return binary { |l, r| l * r } end
def div; return binary { |l, r| l / r } end

private

def unary(&op)
if @stack.size < 1
raise "Insufficient elements on stack for operation."
end

return push( op.call( @stack.pop ) )
end

def binary(&op)
if @stack.size < 2
raise "Insufficient elements on stack for operation."
end

return push( op.call( *@stack.slice!(-2, 2) ) )
end
end

calc = RPNCalc.new
puts calc.push(3)
puts calc.push(5)
puts calc.add
puts calc.push(2)
puts calc.mul
puts calc.push(2)
puts calc.push(3)
puts calc.add
puts calc.div

__END__

I see 3 as the last output when I expect to see 3.2. Why is that?

James Edward Gray II

P.S. I wanted to collapse those math defs to one line for easy reading
and found the semicolon trick used above. Is that the only way to do
it, or is there a method similar to if ... then ... end?
5 Answers

Jamis Buck

9/29/2004 10:15:00 PM

0

James Edward Gray II wrote:
> Quick question. When I run:

[snip]

> I see 3 as the last output when I expect to see 3.2. Why is that?

If you divide two integers, you get an integer (with Ruby). So instead
of say "2" and "3", say "2.0" and "3.0".

--
Jamis Buck
jgb3@email.byu.edu
http://www.jamisbuck...

gabriele renzi

9/29/2004 10:20:00 PM

0

Jamis Buck ha scritto:

> James Edward Gray II wrote:
>
>> Quick question. When I run:
>
>
> [snip]
>
>> I see 3 as the last output when I expect to see 3.2. Why is that?
>
>
> If you divide two integers, you get an integer (with Ruby). So instead
> of say "2" and "3", say "2.0" and "3.0".

or do:
require 'mathn'

to get magic integration of every numerical thingy:
>> 1/2
=> 1/2

Mark Hubbart

9/29/2004 10:28:00 PM

0


On Sep 29, 2004, at 3:08 PM, James Edward Gray II wrote:

> Quick question. When I run:
>
> <snip>
>
> James Edward Gray II
>
> P.S. I wanted to collapse those math defs to one line for easy
> reading and found the semicolon trick used above. Is that the only
> way to do it, or is there a method similar to if ... then ... end?

The other questions were answered pretty well, so I'll just answer the
last one. You can either use the semicolon, or specify an empty
parameter list:

def add; return binary { |l, r| l + r } end
def add() return binary { |l, r| l + r } end

The empty parameter list looks cleaner, imho, though it *is* one
character longer.

cheers,
Mark

James Gray

9/29/2004 10:54:00 PM

0

On Sep 29, 2004, at 5:27 PM, Mark Hubbart wrote:

> The other questions were answered pretty well, so I'll just answer the
> last one. You can either use the semicolon, or specify an empty
> parameter list:
>
> def add; return binary { |l, r| l + r } end
> def add() return binary { |l, r| l + r } end
>
> The empty parameter list looks cleaner, imho, though it *is* one
> character longer.

Thanks to all for the help.

James Edward Gray II

Gavin Kistner

9/30/2004 12:58:00 AM

0

On Sep 29, 2004, at 4:15 PM, Jamis Buck wrote:
>> I see 3 as the last output when I expect to see 3.2. Why is that?
> If you divide two integers, you get an integer (with Ruby). So instead
> of say "2" and "3", say "2.0" and "3.0".

Or:
def div; binary { |l, r| l.to_f / r } end