Re: lexical mumblings

[Miles Bader]

address@hidden (Gerd Moellmann) writes:
> > Of course, the byte-compiler would know that lambda expressions only
> > used by `well-behaved' functions like mapcar can't escape, and so
> > can use the stack-allocated variable-frame.
> 
> Just a nit
> 
> (defvar g)
> (defun f (x) (mapcar (setq g (lambda (e) x)) '(1 2)))
> (f 2)
> (funcall g 1) => ?

In this case, the lambda isn't passed to mapcar, it's passed to setq.

So the compiler would generate a heap-allocated closure, and the
(funcall g 1) would yield `2'.

-Miles
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