please see the code below, why passing a block to a function then call
block.call in the function will get success, but run block.call out of
a function will get failed?
irb(main):023:0* def myhi(&block)
irb(main):024:1> block.call
irb(main):025:1> end
=> nil
Because, as the error message says, this is not valid Ruby syntax. You can use "lambda" or "proc" instead:
irb(main):001:0> lambda { puts "hi" }.call
hi
=> nil
irb(main):002:0>
But this is really useless, i.e. in that case you should simply do "puts 'hi'". The lambda notation really only makes sense if you want to store the block away for later usage. Note, that there is also this syntax (in case you want to structure your code):
begin
puts 'hi'
end
Kind regards
robert
···
On 01/02/2010 11:47 AM, Ruby Newbee wrote:
Hello gurus,
please see the code below, why passing a block to a function then call
block.call in the function will get success, but run block.call out of
a function will get failed?
irb(main):023:0* def myhi(&block)
irb(main):024:1> block.call
irb(main):025:1> end
=> nil
thanks.
but why in the function block.call can be executed?
An incoming code block will be converted into a Proc object and bound
^^^^^^^^^^^^^^^^^^^^^^^^^^^^
to the block variable. You can pass it around to other methods, call it directlyusing call,or yield to it as though you’d never bound it to a variable at all.