Why does it require for the first send() to be None?
You can’t send() a value the first time because the generator did not execute until the point where you have the yield statement, so there is nothing to do with the value.
Here is the relevant paragraph from the pep that introduced the feature of co-routines with generators (http://www.python.org/dev/peps/pep-0342/):
Because generator-iterators begin execution at the top of the
generator’s function body, there is no yield expression to receive
a value when the generator has just been created. Therefore,
calling send() with a non-None argument is prohibited when the
generator iterator has just started, and a TypeError is raised if
this occurs (presumably due to a logic error of some kind). Thus,
before you can communicate with a coroutine you must first call
next() or send(None) to advance its execution to the first yield
expression
A tiny walkthrough:
def coro():
print 'before yield'
a = yield 'the yield value'
b = yield a
print 'done!'
c=coro() # this does not execute the generator, only creates it
# If you use c.send('a value') here it could _not_ do anything with the value
# so it raises an TypeError! Remember, the generator was not executed yet,
# only created, it is like the execution is before the `print 'before yield'`
# This line could be `c.send(None)` too, the `None` needs to be explicit with
# the first use of `send()` to show that you know it is the first iteration
print next(c) # will print 'before yield' then 'the yield value' that was yield
print c.send('first value sent') # will print 'first value sent'
# will print 'done!'
# the string 'the second value sent' is sent but not used and StopIterating will be raised
print c.send('the second value sent')
print c.send('oops') # raises StopIterating