[Python-ideas] A "local" pseudo-function

[Tim Delaney]

​​
On Sat, 28 Apr 2018 at 12:41, Tim Peters <tim.peters at 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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