What do they mean? Why sometimes I should use block and other times &block inside functions that accept blocks?
block
is just a local variable, &block
is a reference to the block passed to the method.
def foo(block = nil)
p block
end
foo # => nil
foo("test") # => test
foo { puts "this block will not be called" } # => nil
def foo(&block)
p block
end
foo # => nil
foo("test") # => ArgumentError: wrong number of arguments (1 for 0)
foo { puts "This block won't get called, but you'll se it referenced as a proc." }
# => #<Proc:0x0000000100124ea8@untitled:20>
You can also use &block
when calling methods to pass a proc as a block to a method, so that you can use procs just as you use blocks.
my_proc = proc {|i| i.upcase }
p ["foo", "bar", "baz"].map(&my_proc)
# => ["FOO", "BAR", "BAZ"]
p ["foo", "bar", "baz"].map(my_proc)
# => ArgumentError: wrong number of arguments (1 for 0)
The variable name block
doesn't mean anything special. You can use &strawberries
if you like, the ampersand is the key here.
You might find this article helpful.
In an argument list, &whatever
takes the block that was passed to the method and wraps it in a Proc object. The Proc is stored in a variable called whatever
(where that can be whatever name you typed after the ampersand, of course — usually it's "block"). After a method call, the &whatever
syntax turns a Proc into a block. So if you define a method like so:
def thing(&block)
thing2 &block
end
You're defining a method that takes a block and then calls another method with that block.
If you don't set the & before block, Ruby won't recognize it's relationship to the "block" you pass to the function. Here some examples.
def f(x, block); end
f(3) { 2+2 } # gives an error, because "block" is a
# regular second argument (which is missing)
def g(x, &block); end
g(3) { 2+2 } # legal
def h(x); end
h(3) { 2+2 } # legal
For later use in a function:
def x(&block) # x is a 0 param function
y(block) # y is a 1 param function (taking one "Proc")
z(&block) # z is a 0 param function (like x) with the block x received
end
So, if you call z(&block)
it's (nearly!!) the same as calling z { yield }
: You just pass the block to the next function.