Printing of kw-args with bang needs a space

[mauro3]

Keyword arguments are usually written without space: fn(;a=1) = a^2. However if the kwarg ends in a bang this does not work:

julia> f(;fn!=x->x[1]=1) = 1
ERROR: syntax: invalid keyword argument syntax "(fn != x -> begin
    # REPL[24], line 1
    x[1] = 1
end)"
Stacktrace:
 [1] top-level scope at REPL[24]:1

and a space is needed. Could no-space be supported?

This would then also be consistent with the current printing of the Expr:

julia> Expr(:kw, :fn!, :a)
:(fn!=a)

[JeffBezanson]

No, I really think it is too confusing to disambiguate fn!=x differently in different contexts. I would say the expression printing is a bug here, since it won't be parsed back correctly.

[mauro3]

Fair enough. I changed the title accordingly.

[ararslan]

TIL this works

julia> f(; (x!)=1) = x! + 1
f (generic function with 1 method)

julia> f()
2

We could just paren-wrap keyword arguments that end with a bang.

[JeffBezanson]

I think the best thing to do is just put spaces around the = like we do for assignment expressions.

[ararslan]

One of the nice things about omitting spaces around = for keyword arguments is that it disambiguates them from assignments. I would personally consider that a regression.

[JeffBezanson]

It doesn't disambiguate them to the parser though.

[JeffBezanson]

In fact, the more comprehensive fix here would be to print kw expressions outside a call as Expr(:kw, ...) and = expressions inside a call as Expr(:(=), ...).

[ararslan]

That makes sense generally speaking, but printing as Expr(:(=), ) loses the visual disambiguation for the user from actual assignments though, since Expr(:(=), :a, :b) prints as :(a = b), so you'd still end up with confusing-looking printing inside of call expressions. That's why I suggest wrapping with parentheses if endswith(a, '!') as part of printing of :kw expressions.

[ararslan]

Or using var"a!".

[JeffBezanson]

IMO the value of printing expressions so that they can be parsed back accurately is worth slightly uglier output in some corner cases.

[ararslan]

I agree with that in general, but printing as Expr(:(=), ) produces uglier output in all cases for the benefit of corner cases.

[JeffBezanson]

No, it's that f(a = 1) parses as Expr(:call, :f, Expr(:kw, :a, 1)). So if you actually have Expr(:call, :f, Expr(:(=), :a, 1)) it should be printed as f($(Expr(:(=), :a, 1)). And the reverse outside a call argument list.