# What is happening with this Unary minus?

LOOPER = 6

-(LOOPER).upto(LOOPER) {|i|
puts i }

I get one line of output: 6

However, I get 13 lines of output here:

(-LOOPER).upto(LOOPER) {|i|
puts i }

What is happening with the unary minus on the first example? I expected
to get identical output.

Todd

Stefano C. wrote:

Alle lunedÃ¬ 27 agosto 2007, Todd B. ha scritto:

I’m not completely sure, but I think the difference arises because of
operator
precedence. The first expression is interpreted as

Stefano

Hi Stefano. I see that now. I did this:

result = -(LOOPER)…
puts result

and I see the minus applied now. Thanks!

Todd

Alle lunedÃ¬ 27 agosto 2007, Todd B. ha scritto:

puts i }

What is happening with the unary minus on the first example? I expected
to get identical output.

Todd

I’m not completely sure, but I think the difference arises because of
operator
precedence. The first expression is interpreted as

-(LOOPER.upto(LOOPER){|i| puts i})

Since the lower and upper bounds are equal, the iteration is performed
only
one time. The - is then applied to the return value of upto (the
i.e LOOPER). Indeed, if you try your code in irb, you’ll see that the
value
of the expression is -6.

In the second case, using brackets you tell the interpreter that the
upto
method should not be called on LOOPER, but on (-LOOPER), that is on -6.

I hope this helps

Stefano

Stefano C. wrote:

precedence. The first expression is interpreted as

I hope this helps

Stefano

This is correct. On the first example, the loop is
(LOOPER).upto(LOOPER) {|i| puts i } added post minus operator