Can anyone explain to me why
"1".to_i*2 is equal to 2
and
"1".to_i *2 is equale to 1?
Thanks
Giovanni
Can anyone explain to me why
"1".to_i*2 is equal to 2
this is ("1".to_i) * ( 2 )
and
"1".to_i *2 is equale to 1?
this is "1".to_i(*2)
Thanks
Non c'è di che.
Giovanni
Robert
···
On 9/30/06, Giovanni Intini <intinig@gmail.com> wrote:
--
Deux choses sont infinies : l'univers et la bêtise humaine ; en ce qui
concerne l'univers, je n'en ai pas acquis la certitude absolue.
- Albert Einstein
Giovanni Intini wrote:
Can anyone explain to me why
"1".to_i*2 is equal to 2
and
"1".to_i *2 is equale to 1?Thanks
Giovanni
It seems to be interpreting
"1".to_i *2
as
"1".to_i(*2)
or
"1".to_i(2)
which is 1. Remember the * in function calls is the unarray operator. Kinda weird in this case, though.
-Justin
because the second version converts the given string in base 2.
to_i accept an optional parameter which is the base of the integer (by
default 10)
http://www.ruby-doc.org/core/classes/String.html#M001463
'1'.to_i*2 calls the method * applied to the value returned by to_i
'1'.to_i *2 is like '1'.to_i(*2)
anyway a more common way to write the expression is
'1'.to_i * 2
Paolo
···
On 30/09/06, Giovanni Intini <intinig@gmail.com> wrote:
Can anyone explain to me why
"1".to_i*2 is equal to 2
and
"1".to_i *2 is equale to 1?Thanks
Giovanni
fr Giovanni:
# "1".to_i*2 is equal to 2
# and
# "1".to_i *2 is equale to 1?
careful w #to_i. It's a powerful converter method that accepts a parameter for the needed conversion.
irb(main):021:0> "111".to_i 2 # base 2 conversion
=> 7
irb(main):022:0> "111".to_i * 2 # base 10 (default) conversion mult by 2
=> 222
irb(main):023:0> "111".to_i 2 # base 2
=> 7
irb(main):024:0> "111".to_i(2) # base 2
=> 7
irb(main):025:0> "111".to_i(*2) # base 2
=> 7
irb(main):026:0> "111".to_i * 2 # base 10 times 2
=> 222
irb(main):027:0> "111".to_i*2 # base 10 times 2
=> 222
irb(main):028:0> "111".to_i *2 # base 2
=> 7
irb(main):029:0> "111".to_i 3 # base 3
=> 13
C:\family\ruby\outlook>ri -T String#to_i
------------------------------------------------------------ String#to_i
str.to_i(base=10) => integer
···
------------------------------------------------------------------------
Returns the result of interpreting leading characters in _str_ as
an integer base _base_ (2, 8, 10, or 16). Extraneous characters
past the end of a valid number are ignored. If there is not a valid
number at the start of _str_, +0+ is returned. This method never
raises an exception.
"12345".to_i #=> 12345
"99 red balloons".to_i #=> 99
"0a".to_i #=> 0
"0a".to_i(16) #=> 10
"hello".to_i #=> 0
"1100101".to_i(2) #=> 101
"1100101".to_i(8) #=> 294977
"1100101".to_i(10) #=> 1100101
"1100101".to_i(16) #=> 17826049
btw, i'm dumb today in windows. i can't run "ri -T String#to_i" in irb. Tips pls.
C:\family\ruby\outlook>ruby -v
ruby 1.8.5 (2006-08-25) [i386-mswin32]
C:\family\ruby\outlook>ver
Microsoft Windows XP [Version 5.1.2600]
kind regards -botp
# Thanks
# Giovanni
This is why you should run with -w as often as possible:
$ ruby -w
"1".to_i *2
-:1: warning: `*' interpreted as argument prefix
···
On Sep 29, 2006, at 4:47 PM, Giovanni Intini wrote:
Can anyone explain to me why
"1".to_i*2 is equal to 2
and
"1".to_i *2 is equale to 1?
--
Eric Hodel - drbrain@segment7.net - http://blog.segment7.net
This implementation is HODEL-HASH-9600 compliant
Collins, Justin wrote:
Remember the * in function calls is the unarray operator. Kinda weird in this case, though.
Reminds me that for Ruby 2, the splat operator was supposed to be a
little more predictable. Anything of that in Ruby 1.9? Can't recall
anything on the change sumamry page about that. This sounds like a
rather nasty gotcha to the people that are lazy about the
one-space-around-operators style convention.
David Vallner