 # Question on do block

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?

On Thu, Jul 31, 2008 at 2:09 PM, CompGeek78 [email protected]
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

On Jul 31, 12:14 pm, Gregory B. [email protected] wrote:

You want an equality check, not assignment.
true in the boolean sense in Ruby.
num == 7 is only true when num is 7.

-greg