iu2 wrote:
On Jul 29, 9:36 am, Duncan Booth <[EMAIL PROTECTED]> wrote:
Your example becomes:
def dotimes(n, callable):
for i in range(n): callable(i)
def block(i):
for j in range(i):
print j,
print
dotimes(5, block)
0
0 1
0 1 2
0 1 2 3
The only thing you asked for that this doesn't do is make 'block'
anonymous, but actually that is a good thing.
Ok, I've got 2 questions about it:
1. Can you please explain why it is a good thing?
All functions defined with lambda expressions get the pseudoname
'<lambda>'. All functions defined with def get the name you give it,
which is typically unique within some scope. The representation of a
function, whether intentionally printed or printed as part of a
traceback, is more meaningful with a specific name than a general name.
2. Will it be possible in Python 3.0 to do the following:
def dotimes(n, callable):
for i in range(n): callable()
def block():
nonlocal i
for j in range(i):
print j,
print
If you indent block so it is a nested function, yes. But the nonlocal
declaration is not needed unless you rebind 'i' from within the nested
function, which block does not do.
tjr
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