Mixing
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 :(:($$ex)) does not do the job.
What is a cleaner way to do this?
julia-lang metaprogramming
asked Nov 20 '18 at 21:24
keornkeorn
132
add a comment |
I have a variable ex that represents an Expr, I want to have a function exprwrap that creates an Expr from it which when evaluated is equal to ex.
Currently I implement it as follows:
ex = :(my + expr)
"Make an expression that when evaled returns the input ex."
function exprwrap(ex::Expr)
ret = :(:(du + mmy))
ret.args[1] = ex
ret
end
eval(exprwrap(ex)) == ex
Keep in mind that my and expr are not defined so :(:($$ex)) does not do the job.
What is a cleaner way to do this?
julia-lang metaprogramming
asked Nov 20 '18 at 21:24
keornkeorn
132
add a comment |
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 :(:($$ex)) does not do the job.
What is a cleaner way to do this?
julia-lang metaprogramming
asked Nov 20 '18 at 21:24
keornkeorn
132
I have a variable ex that represents an Expr, I want to have a function exprwrap that creates an Expr from it which when evaluated is equal to ex.
Currently I implement it as follows:
ex = :(my + expr)
"Make an expression that when evaled returns the input ex."
function exprwrap(ex::Expr)
ret = :(:(du + mmy))
ret.args[1] = ex
ret
end
eval(exprwrap(ex)) == ex
Keep in mind that my and expr are not defined so :(:($$ex)) does not do the job.
What is a cleaner way to do this?
julia-lang metaprogramming
julia-lang metaprogramming
asked Nov 20 '18 at 21:24
keornkeorn
132
asked Nov 20 '18 at 21:24
keornkeorn
132
asked Nov 20 '18 at 21:24
keornkeorn
132
asked Nov 20 '18 at 21:24
keornkeorn
132
asked Nov 20 '18 at 21:24
keornkeorn
132
132
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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("($ex;)") == Expr(:block, ex).
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
2
There is alsoMeta.quotwhich is evaluates toExpr(:quote, x)too.
– 张实唯
Nov 23 '18 at 2:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can write:
Expr(:quote, x)
or
Expr(:block, ex)
or
:($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).
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
2
There is alsoMeta.quotwhich is evaluates toExpr(:quote, x)too.
– 张实唯
Nov 23 '18 at 2:28
add a comment |
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("($ex;)") == Expr(:block, ex).
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
2
There is alsoMeta.quotwhich is evaluates toExpr(:quote, x)too.
– 张实唯
Nov 23 '18 at 2:28
add a comment |
You can write:
Expr(:quote, x)
or
Expr(:block, ex)
or
:($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).
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
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).
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
edited Nov 20 '18 at 21:51
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
answered Nov 20 '18 at 21:37
Bogumił KamińskiBogumił Kamiński
12.8k11220
12.8k11220
2
There is alsoMeta.quotwhich is evaluates toExpr(:quote, x)too.
– 张实唯
Nov 23 '18 at 2:28
add a comment |
2
There is alsoMeta.quotwhich is evaluates toExpr(:quote, x)too.
– 张实唯
Nov 23 '18 at 2:28
2
2
There is also
Meta.quot which is evaluates to Expr(:quote, x) too.– 张实唯
Nov 23 '18 at 2:28
There is also
Meta.quot which is evaluates to Expr(:quote, x) too.– 张实唯
Nov 23 '18 at 2:28
add a comment |
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