Re: [Haskell-cafe] field record update syntax

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)}