How would function calls be differentiated from non-function calls?

[mybearworld]

Consider this:

u foo => bar!
u bar => foo!

Or more verbose:

function foo () => bar()!
function bar () => foo()!

Is it recursive, or will it just output each other's value? How would you achieve the other thing?

[treeplate]

i think line 1 would turn bar() into a string because bar isn't defined yet

[vivaansinghvi07]

not necessarily, depending on the implementation, the program will check if bar is in the namespace only when foo is called

[circle-gon]

Very good point. From the language author's perspective, there isn't really anything we can do. From the programmer, it could be mitigated by inserting a undefined argument, so the interpreter can look at it correctly.

[vivaansinghvi07]

That’s actually exactly how I do it. When running a through the code, any set of () is replaced with an empty Name token, which is interpreted as a DreamberdSpecialBlankValue and ignored when the function itself is called

[circle-gon]

I thought about that, but I thought it would violate the spec (since they aren't always replaced as whitespace), and that it would make parentheses actually have a function instead of being useless (also violating the spec).

[vivaansinghvi07]

That’s true, but I figured that violation is worth having to enforce putting at least one argument in every function call

[TodePond]

This would cause a runtime stack overflow crash.

function foo () => bar()!
function bar () => foo()!

foo()!

As would this:

u foo => bar!
u bar => foo!

foo!

[circle-gon]

So passing functions around by value no longer works? Since it seems like that any mention of a function will automatically invoke it.

[mybearworld]

Hmm, you could probably work around that with c() => bar. Thank you!

[TodePond]

So passing functions around by value no longer works? Since it seems like that any mention of a function will automatically invoke it.

Only for zero-argument functions.