You can do this just fine, but you have to be explicit about what variables you want to pass in, e.g.
let x=2 exp=:(x+1) eval(:(let x = $x; $exp; end)) end If you want to call the expression with multiple inputs, wrap it in a function: let x=2 exp=:(x+1) f = eval(:(x -> $exp)) f(x) end On 11 December 2014 at 06:32, Jameson Nash <vtjn...@gmail.com> wrote: > I'm not quite sure what a genetic program of that sort would look like. I > would be interested to hear if you get something out of it. > > Another alternative is to use a module as the environment: > > module MyEnv > end > eval(MyEnv, :(code block)) > > This is (roughly) how the REPL is implemented to work. > > On Thu Dec 11 2014 at 1:26:57 AM Michael Mayo <mm...@waikato.ac.nz> wrote: > >> Thanks, but its not quite what I'm looking for. I want to be able to edit >> the Expr tree and then evaluate different expressions using variables >> defined in the local scope,not the global scope (e.g. for genetic >> programming, where random changes to an expression are repeatedly evaluated >> to find the best one). Using anonymous functions could work but modifying >> the .code property of an anonymous function looks much more complex than >> modifying the Expr types. >> >> Anyway thanks for your answer, maybe your suggestion is the only possible >> way to achieve this! >> >> Mike >> >> >> On Thursday, December 11, 2014 6:56:15 PM UTC+13, Jameson wrote: >> >>> eval, by design, doesn't work that way. there are just too many better >>> alternatives. typically, an anonymous function / lambda is the best and >>> most direct replacement: >>> >>> let x=2 >>> println(x) # Line 1 >>> exp = () -> x+1 >>> println(exp()) # Line 2 >>> end >>> >>> >>> On Wed Dec 10 2014 at 10:43:00 PM Michael Mayo <mm...@waikato.ac.nz> >>> wrote: >>> >>>> Hi folks, >>>> >>>> I have the following code fragment: >>>> >>>> x=1 >>>> let x=2 >>>> println(x) # Line 1 >>>> exp=:(x+1) >>>> println(eval(exp)) # Line 2 >>>> end >>>> >>>> It contains two variables both named x, one inside the scope defined by >>>> let, and one at global scope. >>>> >>>> If I run this code the output is: >>>> 2 >>>> 2 >>>> >>>> This indicates that (i) that line 1 is using the local version of x, >>>> and (ii) that line 2 is using the global version of x. >>>> >>>> If I remove this global x I now get an error because eval() is looking >>>> for the global x which no longer exists: >>>> >>>> let x=2 >>>> println(x) # Line 1 >>>> exp=:(x+1) >>>> println(eval(exp)) # Line 2 >>>> end >>>> >>>> 2 >>>> >>>> ERROR: x not defined >>>> >>>> >>>> My question: when evaluating an expression using eval() such as line 2, >>>> how can I force Julia to use the local (not global) version of x and thus >>>> avoid this error? >>>> >>>> >>>> Thanks >>>> >>>> Mike >>>> >>>
