Variable / Method Ambiguity

[Summercool Summercool]

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?

 

[Michael Linfield]

So is a Ruby program interpreted or compiled?

interpreted.

 

[David A. Black]

Hi --

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.

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.


David

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?

--
* Books:
RAILS ROUTING (new! http://www.awprofessional.com/title/0321509242)
RUBY FOR RAILS (http://www.manning.com/black)
* Ruby/Rails training
& consulting: Ruby Power and Light, LLC (http://www.rubypal.com)

 

[Summercool Summercool]

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?

 

[Daniel DeLorme]

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]

Hi --

On Mon, 17 Sep 2007, Summercool Summercool 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 this is lexical scoping, then?

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


David

-jf

 

[Gary Wright]

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 --]

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.

 

[7stud --]

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.

 

[Florian Frank]

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"else
a = 1end
end

In your example:

for i in 1..2
if i == 1
x = 1

print "x=", x, "\n"

else
print "x=", x, "\n"

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.

 

[Peña, Botp]

RnJvbTogN3N0dWQgLS0gW21haWx0bzpkb2xndW5AZXhjaXRlLmNvbV0gDQojIEluIHJ1YnksIGZv
ciBsb29wcyBhbmQgaWYgY2xhdXNlcyBkb24ndCBjcmVhdGUgYSBzY29wZS4NCg0KaXJiKG1haW4p
OjAwMTowPiBpZiBmYWxzZQ0KaXJiKG1haW4pOjAwMjoxPiAgICB4DQppcmIobWFpbik6MDAzOjE+
IGVuZA0KPT4gbmlsDQppcmIobWFpbik6MDA0OjA+IHgNCk5hbWVFcnJvcjogdW5kZWZpbmVkIGxv
Y2FsIHZhcmlhYmxlIG9yIG1ldGhvZCBgeCcgZm9yIG1haW46T2JqZWN0DQogICAgICAgIGZyb20g
KGlyYik6NA0KaXJiKG1haW4pOjAwNTowPiBpZiBmYWxzZQ0KaXJiKG1haW4pOjAwNjoxPiAgICB4
ID0gNQ0KaXJiKG1haW4pOjAwNzoxPiBlbmQNCj0+IG5pbA0KaXJiKG1haW4pOjAwODowPiB4DQo9
PiBuaWwNCg0KdGhleSBjYW4gaW50cm9kdWNlIGEgdmFyICh1bmRlciB0aGUgZXhpc3RpbmcgbG9j
YWwgc2NvcGUgb2YgY291cnNlKS4NCmkgdGhpbmsgaXRzIGluIHRoZSBhc3NpZ25tZW50IGJlaGF2
aW91ciBpZiB5b3UgbG9vayBhdCBpdC4gIHBhcnNlciBzZWVzIGFzc2lnbm1lbnQgZXZlbiBpZiBj
b2RlIGlzIG5vdCBleGVjdXRlZC4NCnJ1YnkganVzdCB0cnkgdG8gYmUgZnJpZW5kbHksIGJ1dCB5
b3UgY2FuIGJlIHNsb3BweSBpZiB5b3Ugd2FudCB0byBhbmQgcnVieSB3aWxsIG5vdCBnaXZlIHlv
dSBhbiBlcnJvciA6KQ0KYWdhaW4sIGkgdGhpbmsgdGhpcyBpcyBhIGZhcS4NCg0Ka2luZCByZWdh
cmRzIC1ib3RwDQo=

 

[Vasco Andrade e Silva]

Hi,

David A. Black wrote:

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.

Please, look at this 2 examples of "parse-time":
1)

pp ParseTree.translate('bar = 3

define_methodfoo) do
bar
end')
[:block,
[:lasgn, :bar, [:lit, 3]],
[:iter, [:fcall, :define_method, [:array, [:lit, :foo]]], nil, [:lvar,
:bar]]]

2)

pp ParseTree.translate('bar = 3

def foo
bar
end')
[:block,
[:lasgn, :bar, [:lit, 3]],
[:defn, :foo, [:scope, [:block, [:args], [:vcall, :bar]]]]]

***
In 1) we see a :lvar node, whereas in 2) we see a :vcall. From my
interpretation of p.329 of pickaxe2 i do not agree with the 2nd case
(but probably it's my problem - comment please). In the 2nd case, IMHO,
we should see a lvar node.
This may seem a detail, but i was bitten by that a few minutes ago. Let
me show a simplified example:

bar = "test"
c = Class.new
c.class_eval do
def foo
bar
end
end

c.new.foo

NameError: undefined local variable or method `bar'
# i was expecting to get "test"

The (hack) "solution" (it's a hack because IMHO i shouldn't have to do
that):
bar = "test"
c = Class.new
c.class_eval do
define_methodfoo) do
bar
end
end

c.new.foo

"test"

Note: only one more thing, since ruby 1.8 have a "problem" with block
(it can't get other blocks - &block in argument) this could be a
problem. I know you can hack these thing through a method - but that's
another hack ... too much hacks for me

I would appreciate very much all your comments about the current
behavior of ruby (1.8).

Regards,
Vasco Andrade e Silva