Asp Forum - Variable / Method Ambiguity

SpringFlowers AutumnMoon

9/17/2007 1:19:00 AM

the following program for showing Variable / Method Ambiguity

def a
print "Function 'a' called\n"
99
end

for i in 1..2
if i == 2
print "a=", a, "\n"
else
a = 1
print "a=", a, "\n"
end
end

will produce the result:

a=1
Function 'a' called
a=99

(as described in the PickAx2 book, p. 329)

So at first I thought a Ruby program is interpreted? If interpreted,
then the interpreter will see the "a = 1" at first as treat "a" as a
variable, and then the second time it sees "a", the interpreter should
treat it as a variable again, not as a method. So does that mean a
Ruby program is not interpreted but compiled into some bytecode first?
But then, sometimes I run a Ruby program and then it can run all the way
until it see an "undefined local variable"... so that means it probably
is not compiled... or else it would have stopped without running
anything at all.

So is a Ruby program interpreted or compiled?
--
Posted via http://www.ruby-....

10 Answers

Michael Linfield

9/17/2007 7:00:00 AM

> So is a Ruby program interpreted or compiled?

interpreted.
--
Posted via http://www.ruby-....

David A. Black

9/17/2007 7:49:00 AM

SpringFlowers AutumnMoon

9/17/2007 9:54:00 AM

On Sep 17, 12:49 am, "David A. Black" <dbl...@rubypal.com> wrote:
>
> No, because there's no such variable in scope. The second time through
> the loop, there's no assignment to a, and the first variable a has
> gone out of scope. So the only thing that "a" could mean, outside of
> an assignment, is the method a.

wow... i thought the "a" variable is kind of like in C or Python:

for (i = 1; i <= 2; i++) {
int a;

if (i==2) {
...
} else {
...
}

}

so that the "a" exists all inside the for loop... maybe that's not
right.

now, we can view the variable "a" as existing in the else block but no
where else in the original Ruby code?

--
Posted via http://www.ruby-....

Daniel DeLorme

9/17/2007 2:30:00 PM

Summercool Summercool wrote:
> So at first I thought a Ruby program is interpreted? If interpreted,
> then the interpreter will see the "a = 1" at first as treat "a" as a
> variable, and then the second time it sees "a", the interpreter should
> treat it as a variable again, not as a method. So does that mean a
> Ruby program is not interpreted but compiled into some bytecode first?
> But then, sometimes I run a Ruby program and then it can run all the way
> until it see an "undefined local variable"... so that means it probably
> is not compiled... or else it would have stopped without running
> anything at all.
>
> So is a Ruby program interpreted or compiled?

It is interpreted but before being interpreted it is parsed. And whether
a symbol is a variable or not is determined at parse time: when the
parser sees the assignment "a = 1", it assumes that 'a' is a variable
for all subsequent *lines* of the source code, until the end of the
method or block where the assignment occured. It is has nothing to do
with the order in which the code is executed.

Also you'll notice that the "undefined local variable" message is
actually "undefined local variable or method", because ruby can't know
if the unknown symbol was supposed to be a method or a variable.

Daniel

Jeffrey 'jf' Lim

9/17/2007 4:08:00 PM

On 9/17/07, David A. Black <dblack@rubypal.com> wrote:
> Hi --
>
> On Mon, 17 Sep 2007, Summercool Summercool wrote:
>
> <snip>
> >
> > (as described in the PickAx2 book, p. 329)
> >
> > So at first I thought a Ruby program is interpreted? If interpreted,
> > then the interpreter will see the "a = 1" at first as treat "a" as a
> > variable, and then the second time it sees "a", the interpreter should
> > treat it as a variable again, not as a method.
>
> No, because there's no such variable in scope. The second time through
> the loop, there's no assignment to a, and the first variable a has
> gone out of scope.

so this is lexical scoping, then?

> So the only thing that "a" could mean, outside of
> an assignment, is the method a.
>
>
> David
>

-jf

--
In the meantime, here is your PSA:
"It's so hard to write a graphics driver that open-sourcing it would not help."
-- Andrew Fear, Software Product Manager, NVIDIA Corporation
http://kerneltrap.org...

Gary Wright

9/18/2007 7:07:00 PM

On Sep 17, 2007, at 12:08 PM, Jeffrey 'jf' Lim wrote:

> On 9/17/07, David A. Black <dblack@rubypal.com> wrote:
>> Hi --
>>
>> On Mon, 17 Sep 2007, Summercool Summercool wrote:
>>
>> <snip>
>>>
>>> (as described in the PickAx2 book, p. 329)
>>>
>>> So at first I thought a Ruby program is interpreted? If
>>> interpreted,
>>> then the interpreter will see the "a = 1" at first as treat "a" as a
>>> variable, and then the second time it sees "a", the interpreter
>>> should
>>> treat it as a variable again, not as a method.
>>
>> No, because there's no such variable in scope. The second time
>> through
>> the loop, there's no assignment to a, and the first variable a has
>> gone out of scope.
>
> so this is lexical scoping, then?

No, it is just that Ruby has a single-pass parser. The parser needs
to decide if an identifier is a local variable or a zero-argument method
before it has seen the entire text of the file (no look-ahead). Until
the parser sees an assignment of the form "a = ..." it assumes
that 'a' is a method call.

So if/then/else statements do *not* create new lexical scopes.

Gary Wright

7stud 7stud

9/18/2007 8:12:00 PM

David A. Black wrote:
> No, because there's no such variable in scope. The second time through
> the loop, there's no assignment to a, and the first variable a has
> gone out of scope. So the only thing that "a" could mean, outside of
> an assignment, is the method a.
>

That doesn't seem to bear out. If you switch the if statement around so
that the assignment happens first:

def x
print "Function 'x' called\n"
99
end

for i in 1..2
if i == 1
x = 10
print "x=", x, "\n"
else
print "x=", x, "\n"
end
end

the output is:

x=10
x=10

The second time through that loop, there is no assignment to x and the
previous x has gone out of scope, yet the else clause's use of x does
not result in a method call.

As p. 329 in pickaxe2 carefully explains, it is a ruby parsing rule that
determines what x means. As I read the rule, and as Daniel explains,
and as the two examples show(the 'a' example and the 'x' example), the
rule is: if there is any code that has x=val on a line before the use of
x, then x will be interpreted as a variable. If there is no assignment
to x on a line higher in the code, then x will be interpreted as a
method call.

--
Posted via http://www.ruby-....

7stud 7stud

9/18/2007 8:21:00 PM

7stud -- wrote:
>David Black wrote:
>The second time through
>the loop, there's no assignment to a, and the first variable a has
>gone out of scope. >

Wait a minute. 'a' has gone out of scope?

for i in 1..2
if i == 1
a = 10
print "a=", a, "\n"
else
puts a
end
end

puts "*#{a}"

puts a

--output:--
a=10
10
*10

In ruby, for loops and if clauses don't create a scope.
--
Posted via http://www.ruby-....

Florian Frank

9/18/2007 8:37:00 PM

7stud -- wrote:
> That doesn't seem to bear out. If you switch the if statement around so
> that the assignment happens first:
>
> def x
> print "Function 'x' called\n"
> 99
> end
>
>
The difference between the both examples is the following:

Ruby parses this:

for i in 1..2
if i == 2
print "a=", a, "\n"
>> Here Ruby decides, that "a" must mean the a method and creates the call-method node in its tree.
else
a = 1
>> Here Ruby decides, a is a local variable.
print "a=", a, "\n"
>> Here Ruby creates a evaluate-local-variable node in its tree.
end
end

In your example:

> for i in 1..2
> if i == 1
> x = 1
>
>> Here Ruby decides, "x" is a local variable.

> print "x=", x, "\n"
>
>> Here Ruby creates a evaluate-local-variable node in its tree.

> else
> print "x=", x, "\n"
>
>> Sticks to its former decision, that "x" is local variable, and
creates a evaluate-local-variable node in its tree.
> end
> end
>
This is an interaction between parsing and execution or ruby code, and
it can be a bit surprising. The later execution (== comparisons, etc.)
doesn't alter the behavior of the parser, which does its jobs earlier.

--
Florian Frank

Peña, Botp

9/19/2007 2:51:00 AM

From: 7stud -- [mailto:dolgun@excite.com]
# In ruby, for loops and if clauses don't create a scope.

irb(main):001:0> if false
irb(main):002:1> x
irb(main):003:1> end
=> nil
irb(main):004:0> x
NameError: undefined local variable or method `x' for main:Object
from (irb):4
irb(main):005:0> if false
irb(main):006:1> x = 5
irb(main):007:1> end
=> nil
irb(main):008:0> x
=> nil

they can introduce a var (under the existing local scope of course).
i think its in the assignment behaviour if you look at it. parser sees assignment even if code is not executed.
ruby just try to be friendly, but you can be sloppy if you want to and ruby will not give you an error :)
again, i think this is a faq.

kind regards -botp

comp.lang.ruby

Variable / Method Ambiguity

SpringFlowers AutumnMoon

Michael Linfield

David A. Black

SpringFlowers AutumnMoon

Daniel DeLorme

Jeffrey 'jf' Lim

Gary Wright

7stud 7stud

7stud 7stud

Florian Frank

Peña, Botp

x Login to ForumsZone