CODEWITHSUNDEEP
juliametaprogramming
keorn
keorn

Reputation: 75

How to create an Expr that evaluates to Expr in Julia?

I have a variable ex that represents an Expr, I want to have a function exprwrap that creates an Expr from it which when evaluated is equal to ex.

Currently I implement it as follows:

ex = :(my + expr)

"Make an expression that when evaled returns the input ex."
function exprwrap(ex::Expr)
  ret = :(:(du + mmy))
  ret.args[1] = ex
  ret
end

eval(exprwrap(ex)) == ex

Keep in mind that my and expr are not defined so :(:($$ex)) does not do the job.

What is a cleaner way to do this?

Upvotes: 2

Views: 343

Answers (1)

Bogumił Kamiński
Bogumił Kamiński

Reputation: 69829

You can write:

Expr(:quote, x)

or

Expr(:block, ex)

or

:($ex;)

Additionally you could do:

Meta.parse(":($ex)")

which is not simple but shows you how Julia parses ex when it appears in the source code and you can see that it is the same as Expr(:quote, ex).

Similarly yo can check that Meta.parse("($ex;)") == Expr(:block, ex).

Upvotes: 3

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