Question on do block

ruby-talk
1

While going through some practice code from a book, I ran into an odd
issue with this do block.

The book put the code this way and it worked fine.

def mtdarry
  10.times do |num|
    square = num * num
    return num, square if num > 7
  end
end

num, square = mtdarry
puts num
puts square

This returns 8 and 64, which makes sense.

The problem I ran into is in changing the > to a =.

def mtdarry
  10.times do |num|
    square = num * num
    return num, square if num = 7
  end
end

num, square = mtdarry
puts num
puts square

At this point, it outputs 7 and 0. Why does it not calculate the value
of square properly?

2

The problem I ran into is in changing the > to a =.

def mtdarry
10.times do |num|
   square = num * num
   return num, square if num = 7
end
end

You want an equality check, not assignment.

a = 1

=> 1

a == 1

=> true

a == 2

=> false

num = 7 is always true, because all values except false and nil are
true in the boolean sense in Ruby.
num == 7 is only true when num is 7.

-greg

···

On Thu, Jul 31, 2008 at 2:09 PM, CompGeek78 <keven.denen@gmail.com> wrote:

3

Oh gads I'm an idiot...thanks.

···

On Jul 31, 12:14 pm, Gregory Brown <gregory.t.br...@gmail.com> wrote:

On Thu, Jul 31, 2008 at 2:09 PM, CompGeek78 <keven.de...@gmail.com> wrote:
> The problem I ran into is in changing the > to a =.

> def mtdarry
> 10.times do |num|
> square = num * num
> return num, square if num = 7
> end
> end

You want an equality check, not assignment.

>> a = 1
=> 1
>> a == 1
=> true
>> a == 2

=> false

num = 7 is always true, because all values except false and nil are
true in the boolean sense in Ruby.
num == 7 is only true when num is 7.

-greg