On Sat, 28 Apr 2018 at 12:41, Tim Peters <tim.pet...@gmail.com> wrote:
My big concern here involves the:
if local(m = re.match(regexp, line)):
print(m.group(0))
example. The entire block needs to be implicitly local for that to work -
what happens if I assign a new name in that block? Also, what happens with:
if local(m = re.match(regexp1, line)) or local(m = re.match(regexp2, line)
):
print(m.group(0))
Would the special-casing of local still apply to the block? Or would you
need to do:
if local(m = re.match(regexp1, line) or re.match(regexp2, line)):
print(m.group(0))
This might just be lack of coffee and sleep talking, but maybe new
"scoping delimiters" could be introduced. Yes - I'm suggesting introducing
curly braces for blocks, but with a limited scope (pun intended). Within a
local {} block statements and expressions are evaluated exactly like they
currently are, including branching statements, optional semi-colons, etc.
The value returned from the block is from an explicit return, or the last
evalauted expression.
a = 1
> b = 2
> c =
>
> local(a=3) * local(b=4)
>
c = local { a=3 } * local { b=4 }
c =
>
> local(a=3
> ,
> b=4
> ,
> a*b)
c =
local
{
a=3
;
b=4
;
a*b
}
c =
local
{
a = 3
b = 4
a * b
}
c = local(a=3,
>
> b=local(a=2, a*a), a*b)
c =
local
{
a = 3
b = local(a=2, a*a)
return a * b
}
>
> r1, r2 = local(D = b**2 - 4*a*c,
> sqrtD = math.sqrt(D),
> twoa = 2*a,
> ((-b + sqrtD)/twoa, (-b - sqrtD)/twoa))
>
r1, r2 = local {
D = b**2 - 4*a*c
sqrtD = math.sqrt(D)
twoa = 2*a
return ((-b + sqrtD)/twoa, (-b - sqrtD)/twoa)
}
> if local(m = re.match(regexp, line)):
> print(m.group(0))
>
if local { m = re.match(regexp, line) }:
print(m.group(0))
And a further implication:
a = lambda a, b: local(c=4, a*b*c)
a = lambda a, b: local {
c = 4
return a * b * c
}
Tim Delaney
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