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Ruby block, procs and instance_eval

I recently tried to do something akin to this:

a = "some string"
b = Proc.new{ upcase }
a.instance_eval b

Which gives the error:

TypeError: can't convert Proc into String

but this works:

def b(&block)
  "some string".instance_eval &block
end

b{ upcase }

A further look with this method:

def b(&block)
  "some string".instance_eval block
end

Yields the same Proc to String error.

So... my understanding of blocks is that they are just procs. But there's obviously something special about having this & ampersand...

Can someone explain this to me? Is it possible to convert a normal proc to be whatever it is that is special about this &block object?

edit

Just figured out my second question, prepend an & to the proc... that was easy, but WHAT is this really doing?

like image 306
brad Avatar asked Jun 16 '11 13:06

brad


3 Answers

All you have to do for your first example to work is this:

>> a.instance_eval &b #=> "SOME STRING"

The reason is that instance_eval needs either a string or a block and the ampersand provides the latter.

like image 186
Michael Kohl Avatar answered Sep 19 '22 08:09

Michael Kohl


The difference is that a.instance_eval b is passing b as a regular argument to instance_eval, whereas a.instance_eval &b is passing it as a block. Those are two different things.

Consider this method call:

obj.foo(bar) do |x| 
  stuff(x) 
end

That invokes the method foo with one regular argument (bar) and one block argument (do |x| stuff(x) end). In the method definition, they're distinguished by prefixing & to the block parameter:

def foo(arg, &block)

And if you wish to pass a variable expression instead of a literal block, that is likewise accomplished by prefixing & to the expression (which should yield a Proc).

If you pass an argument with no &, it goes in the arg slot instead of the block slot. It doesn't matter that the argument happens to be an instance of Proc. The syntax dictates how it is passed and treated by the method.

like image 24
Mark Reed Avatar answered Sep 18 '22 08:09

Mark Reed


It's because instance_eval accepts a string to eval or a block. instance_eval(&block) is passing your block as a block to instance_eval.

like image 22
Samuel Avatar answered Sep 17 '22 08:09

Samuel