# Pattern Match Bug with Equals?

I'm trying to define a pattern matching function:

test[x__?MatchQ[Equal[_, _]]] := 1;


Now I give it a series of inputs, and expect each to output 1:

test[1 == a, 1 == b]
test[1 == a, 1 == a]
test[1 == a, a == 1]
test[1 == a]


...none of which evaluate to 1. I can't see what I've done wrong here. So I try the following:

MatchQ[Equal[_, _]][a == b]


Which gives False. I don't understand why. I look at the FullForm of a==b and it gives back Equal[a,b] which should match with the pattern Equal[_,_].

However, for reasons beyond me, this expression...

MatchQ[Equal[_, __]][a == b]


...evaluates to True.

I have no idea what's going on here or why my test pattern match doesn't work. Can someone please explain what is going on? i.e.

• Why doesn't my test pattern match work as intended?
• How can I pattern match a sequence of equalities a==b, c==d ...?
• Why does MatchQ[Equal[_,__]][a==b] yield True while replacing __ with _ yields false?
• A few things going on here. a) look at Equal[_, _]. It's True. So you'll need a HoldPattern. b) ? is a high-precedence operator so you need ?(MatchQ[HoldPattern[Equal[_, _]]]). Even better though would just be x__Equal. It's most concise and the fastest. Dec 21, 2017 at 5:54
• @b3m2a1 - Thanks! That pretty much answers it. Looks like Mathematica doesn't evaluate _==__ so that explains the behaviour of that pattern match. You can write up a brief answer if you want. Dec 21, 2017 at 5:56
• Feel free to self-answer. I'm checking to see if this is a dupe (but a self-answer won't hurt either way). Dec 21, 2017 at 5:58
• You could also write HoldPattern[_ == _] .. Dec 21, 2017 at 8:52

• ? is a high-precedence operator, and so its argument should be encased in parentheses. i.e. ?(MatchQ[HoldPattern[Equal[_, _]]]) instead of ?MatchQ[HoldPattern[Equal[_, _]]]
• Mathematica is evaluating Equal[_,_] to True before pattern matching! We need to use HoldPattern[Equal[_,_]].
• One more thing to add: test[x__Equal] := 1 will work, too (and probably be faster) Dec 21, 2017 at 6:14
• @b3m2a1 - they have different behaviour, however. If the user enters x==y==z, I don't want this to be pattern matched. Dec 21, 2017 at 7:01
• Ahh. Okay. I assumed you were only working with a==b type expressions. Dec 21, 2017 at 7:02