I do a lot of this soft of thing.
foo {bar = fn $ bar foo
,baz = fn2 $ baz foo
}It would be much nicer if this syntax did the equivalent:
foo {bar \= fn
,baz \= fn2
}
-Alex-
On Thu, 27 Jan 2005, Henning Thielemann wrote:
On Thu, 27 Jan 2005, S. Alexander Jacobson wrote:
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...
If you have only few different record fields you may like to define an update function for each record field.
updateBar fn foo = foo {bar = fn (bar foo)}
______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com _______________________________________________ Haskell mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell
