Re: [RFC] Draft - Closure self reference

On Fri, Nov 20, 2020, at 10:01 AM, David Rodrigues wrote: > > Is it returnType as $thing ? That doesn't read well, > > but I don't see >> us making return type movable. as/use should be placable in either order, >> but maybe we give them required order anyway? > > I don't know if that would be a problem, because today we can have it > "function(): Type use($x)", so "Type use($x)"? I have to agree that it may > sound strange at first, but I also can't imagine many ways of doing things > differently without creating more complexities. > > (1) function as $lambda (... $args): returnType use ($captures...) {} > (2) function (... $args) as $lambda: returnType use ($captures...) {} > (3) function (... $args): returnType as $lambda use ($captures...) {} // > original suggestion > (4) function (... $args): returnType use ($captures...) as $lambda {} > (4.b) function (... $args): returnType as $lambda {} // without use() will > be same as (3) > > Or, define a "inline function" with its own name, visible only within the > scope itself. Similar to how JS works. > > (5) function $lambda(... $args): returnType use ($captures...) {} > > To be honest, as I typed what came to mind, I ended up preferring this last > option. I kind of like that option, too. The main question in my mind is whether that means the same variable is used internally and externally or not. That is, my knee-jerk reaction when reading that was to ask if this would be confusing: $fib = function $f(int $n): int { ... } Viz, do $fib and $f need to match? It looks from the example below like you're saying it's actually: function $fib(int $n): int { ... } Where $fib is then both the internal and external variable name. That seems unexpected. It may be the best way to go about it, but it feels surprising as a user space developer. I do think something like that is probably the best way forward, though. The other question of course is short-lambdas. Would that mean this? fn $fib(int $n): int => match($n) { 0 => 0, 1 => 1, default => $fib(n - 1) + $fib($n - 2), }; If the function is inlined, does that mean the variable name leaks and thus remains in scope (and the closure thus using memory)? array_map(fn $fib(int $n): int => match($n) { 0 => 0, 1 => 1, default => $fib(n - 1) + $fib($n - 2), }, range(1, 6)); print $fib(5); // Does this work? (No you shouldn't write it like that, but since Fibonacci is the example we're working with...) --Larry Garfield