So all the Xs would be in scope at s6. Important point. Thanks, Michael --- On Tue, 8/10/10, Alex Stangl <alex@stangl.us> wrote: From: Alex Stangl <alex@stangl.us> Subject: Re: [Haskell-cafe] Couple of questions about *let* within *do* To: "michael rice" <nowgate@yahoo.com> Cc: "Tillmann Rendel" <rendel@Mathematik.Uni-Marburg.de>, haskell-cafe@haskell.org Date: Tuesday, August 10, 2010, 2:21 PM On Tue, Aug 10, 2010 at 11:01:28AM -0700, michael rice wrote:
Hi all,
Then,
do s1 s2 let x1 = e1 x2 = e2 s3 s4 let x3 = e3 x4 = e4 s5 s6
becomes
do s1 s2 let x1 = e1 x2 = e2 in do s3 s4 let x3 = e3 x4 = e4 in do s5 s6?
do s1 s2 let x1 = e1 x2 = e2 in do s3 s4 let x3 = e3 x4 = e4 in do s5 s6 HTH, Alex