The symbol to proc operator, or not?

Some time ago I used to think that the & unary operator in Ruby could be used to create a Proc object. It turns out I was slightly wrong.

I think I’m not alone here as apparently the &: operator even gets the name of symbol to proc operator. The name is misleading in several ways:

It’s not really an operator, &: is a combination of the & operator followed by a symbol definition. &:upcase is exactly the same as &(:upcase).
You can’t use it to create a Proc object, this is explained this with detail below.

Using it at method invocation (proc to block)

To be precise, the operator does create a Proc object when used with a symbol, but it’s just because Ruby needs to do this internally as part of the process to build the block which will be passed to the method being invoked. As this proc is internal, we can’t access it from our Ruby code.

What the operator really does is to pass its only operand as a block to the method we’re invoking. It converts something to a block.

Take the following snippet for example:

%w(a b c).map(&:upcase)
# => ["A", "B", "C"]

When this is executed, Ruby will pass whatever follows the & operator as the block for the method. But in this case we have a Symbol object.

It’s pretty obvious how to convert a Proc into a block. Both things are almost the same.

It’s not so obvious how to convert a Symbol to a block. Here’s where :to_proc comes in. When the object given to the operator isn’t a proc, :to_proc will be invoked on the parameter to produce a proc that will later be converted to a block.

So, the snippet can be rewritten as the following equivalent:

%w(a b c).map(&(:upcase.to_proc))
# => ["A", "B", "C"]

Notice that we haven’t used the operator to build the proc, but to build the block.

Using it at method definition (block to proc)

Just for completeness, I’ll mention that the same operator can be used in a different context to convert a block to a proc.

In the following example, the operator is used in the params of a method definition:

def repeat_twice(&what)
  what.call
  what.call
end

repeat_twice do
  puts rand(10)
end

The code will show a couple of random numbers on screen, but what’s most important here is that the block defined with a do ... end becomes a proc object named what inside the method.

When the repeat_twice is invoked, we’re sending a block which will be converted to a proc and added to the table of local variables when the method is executed.

In this case we did create a proc, but we didn’t use any Symbol at all.

So what?

In summary, you can’t use the symbol to proc operator to create a proc from a symbol.

This looks like a subtle thing, but hopefully I’ve helped someone to avoid the confusion I had.

I guess block operator is a better name, because what it really does is to convert procs to blocks and vice-versa.

What do you think?