Perhaps comparing ASTs isn't a very common operation, but I came across the following which I found surprising. I think it is even arguable that == should ignore the facts that NaNs are not numerically equal to themselves since it's not really numerical comparison.
julia> :(f($NaN)) == :(f($NaN))
false
julia> isequal(:(f($NaN)), :(f($NaN)))
false
Keno
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It looks like we need an isequal method for Expr that calls isequal recursively.
JeffBezanson
on 31 Mar 2016
👍1
it is even arguable that == should ignore the facts that NaNs are not numerically equal to themselves since it's not really numerical comparison.
NaN literal is not NaN value, yep
wizzard0
on 2 Apr 2016
🎉1
👍1
Very good point. The real solution is probably for == on Exprs to use isequal to compare elements.
JeffBezanson
on 3 Apr 2016
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NaN literal is not NaN value, yep