
27 Jan
2005
27 Jan
'05
12:59 p.m.
I have a lot of code of the form foo {bar = fn $ bar foo} Is there a more concise syntax? I am thinking the record equivalent of C's foo+=5... I imagine there is some operator that does this e.g. foo {bar =* fn} But I don't know what it is... -Alex- ______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com