why do I get a syntax error for
puts 1 if ( 1 == 2
or 1 == 1 )
but there is no syntax error for
puts 1 if ( 1 == 2 or
1 == 1 )i don’t get one with the ruby-1.8.1 that’s distributed in debian’s
sid.
···
On Sun, Jan 18, 2004 at 08:10:01AM +0900, Rick Hu wrote:
why do I get a syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )but there is no syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )
A side-effect of Ruby allowing you to use or not use semicolons as you
please.
The Pickaxe says “Ruby expressions and statements are terminated at the
end of a line unless the statement is obviously incomplete – for example
if the last token on a line is an operator or a comma. A semicolon can be
used to separate multiple expressions on a line. You can also put a
backslash at the end of a line to continue it onto the next.”
···
On Sat, 17 Jan 2004 15:07:57 -0800, Rick Hu wrote:
why do I get a syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )but there is no syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )
rick.hu@gol.com (Rick Hu) wrote in message news:e22cc553.0401171507.a8f41a2@posting.google.com…
why do I get a syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )but there is no syntax error for
puts 1 if ( 1 == 2 or 1 == 1 )
I just found out that the parensis can start a COMPOSITE statement.
This explains why the first case is a syntax error, and why
( boolean
boolean
…
boolean )
returns the value of the last value.
messju mohr wrote:
···
On Sun, Jan 18, 2004 at 08:10:01AM +0900, Rick Hu wrote:
why do I get a syntax error for
puts 1 if ( 1 == 2
or 1 == 1 )but there is no syntax error for
puts 1 if ( 1 == 2 or
1 == 1 )i don’t get one with the ruby-1.8.1 that’s distributed in debian’s
sid.
I do, with Ruby 1.8.1. But the reason (as far as I’ve been able to
ascertain) is that Ruby’s grammar does not allow newlines before operators.
–
Jamis Buck
jgb3@email.byu.edu
ruby -h | ruby -e ‘a=;readlines.join.scan(/-(.)[e|Kk(\S*)|le.l(…)e|#!(\S*)/) {|r| a << r.compact.first };puts “\n>#{a.join(%q/ /)}<\n\n”’
messju mohr wrote:
why do I get a syntax error for
puts 1 if ( 1 == 2
or 1 == 1 )but there is no syntax error for
puts 1 if ( 1 == 2 or
1 == 1 )i don’t get one with the ruby-1.8.1 that’s distributed in debian’s
sid.I do, with Ruby 1.8.1. But the reason (as far as I’ve been able to
ascertain) is that Ruby’s grammar does not allow newlines before operators.
i also get one if i run
puts 1 if ( a == b
or a == c )
maybe just because “or” is an operator and “a” isn’t. i avoid newlines
inside ruby-statements because i don’t know the rules i have to
obey. where are they stated?
···
On Sun, Jan 18, 2004 at 08:37:49AM +0900, Jamis Buck wrote:
On Sun, Jan 18, 2004 at 08:10:01AM +0900, Rick Hu wrote:
–
Jamis Buck
jgb3@email.byu.eduruby -h | ruby -e
‘a=;readlines.join.scan(/-(.)[e|Kk(\S*)|le.l(…)e|#!(\S*)/) {|r| a <<
r.compact.first };puts “\n>#{a.join(%q/ /)}<\n\n”’
messju mohr wrote:
i also get one if i run
puts 1 if ( a == b
or a == c )maybe just because “or” is an operator and “a” isn’t. i avoid newlines
inside ruby-statements because i don’t know the rules i have to
obey. where are they stated?
I don’t know where they are authoritatively stated, except in the Ruby
source code. 
“or” is, indeed, an operator. So are ‘==’, ‘+’, ‘and’,
‘&&’, ‘not’, ‘!’, ‘=~’, and so forth.
···
–
Jamis Buck
jgb3@email.byu.edu
ruby -h | ruby -e ‘a=;readlines.join.scan(/-(.)[e|Kk(\S*)|le.l(…)e|#!(\S*)/) {|r| a << r.compact.first };puts “\n>#{a.join(%q/ /)}<\n\n”’
== and + (among others) are methods, not operators.
Gavin
···
On Sunday, January 18, 2004, 10:59:05 AM, Jamis wrote:
messju mohr wrote:
i also get one if i run
puts 1 if ( a == b
or a == c )maybe just because “or” is an operator and “a” isn’t. i avoid newlines
inside ruby-statements because i don’t know the rules i have to
obey. where are they stated?
I don’t know where they are authoritatively stated, except in the Ruby
source code.
‘&&’, ‘not’, ‘!’, ‘=~’, and so forth.
Thanks for all your responses.
Now try these
puts 1 if ( 1 == 2
1 == 1 )
[ syntax is correct. output 1 ]
puts 1 if ( 1 == 1
1 == 2 )
[ syntax is correct. output nothing ]
Somehow,
( boolean
boolean
boolean
…
boolean )
is a valid syntax and the last boolean value is the value of the expression.
What’s the rationale behind this?
Gavin Sinclair wrote:
== and + (among others) are methods, not operators.
Gavin
You are, of course, correct. However, I still stand by my statement
that == and + are also operators, since they have a precedence and
associativity, which your run-of-the-mill ‘method’ does not have. Also,
it makes it easier to remember where you can and cannot put newlines.
···
–
Jamis Buck
jgb3@email.byu.edu
ruby -h | ruby -e ‘a=;readlines.join.scan(/-(.)[e|Kk(\S*)|le.l(…)e|#!(\S*)/) {|r| a << r.compact.first };puts “\n>#{a.join(%q/ /)}<\n\n”’
No, they’re operators, which happen to result in a method invocation.
The definition of “operator” is syntactic, not semantic, and
the difference is evident in the way they’re used: a + b vs. a.+(b).
I would expect an unclosed parenthesis to be enough to imply
continuation on the next line, and not generate a syntax error.
-Mark
···
On Sun, Jan 18, 2004 at 09:31:27AM +0900, Gavin Sinclair wrote:
== and + (among others) are methods, not operators.
Rick Hu wrote:
Thanks for all your responses.
Now try these
puts 1 if ( 1 == 2
1 == 1 )[ syntax is correct. output 1 ]
puts 1 if ( 1 == 1
1 == 2 )[ syntax is correct. output nothing ]
Somehow,
( boolean
boolean
boolean
…
boolean )
is a valid syntax and the last boolean value is the value of the expression.What’s the rationale behind this?
What’s going on here is:
- Ruby has two valid statement terminators - semicolon and end-of-line
- Expressions can be grouped in parentheses
- The line between statements and expressions is blurred in Ruby
Observe that, since Ruby is expression-oriented,
1 == 2
is valid “stand-alone” line of code. It just doesn’t do anything; it’s
evaluated and thrown away.
Ruby has no real concept of a “Boolean,” by the way. false and nil are
treated as false; everything else is treated as true. The value true
exists for convenience, of course.
Hal