Re: confused about == vs === in this equality
- To: mathgroup at smc.vnet.net
- Subject: [mg103769] Re: confused about == vs === in this equality
- From: Szabolcs Horvát <szhorvat at gmail.com>
- Date: Mon, 5 Oct 2009 13:16:01 -0400 (EDT)
- References: <20091003104738.LCJ3I.416659.imail@eastrmwml34> <email@example.com>
On 2009.10.04. 12:44, Nasser Abbasi wrote: > ?=== > lhs===rhs yields True if the expression lhs is identical to rhs, and yields > False otherwise. > > ?== > lhs==rhs returns True if lhs and rhs are identical. > > But looking at this example: > > a = ComplexInfinity; > If[a == ComplexInfinity, Print["YES"]] > > Expecting it would print "YES", but it does not. it just returns the whole > thing unevaluated? But > > If[a === ComplexInfinity, Print["YES"]] > > does return YES. > > I guess I am a little confused about the "expression" bit in the definition. > > So, when using the 3"=", it is looking at the _value_ of the expression, but > when using the 2"=", it is looking at the expression _as it is_, i.e. > without evaluating it? Is this the difference? I've always used the 2"=" > for equality, now I have to be more careful which to use. > === tests for structural equivalence, without taking into account any mathematical meaning. It yields True if and only if the FullForm of the lhs and rhs look identical. == tests for mathematical equivalence, AND is also used when writing equations. ComplexInfinity == ComplexInfinity won't give true because the phase of ComplexInfinity is undetermined. Comparing ComplexInfinity to a finite number will always give False. None of the two operators have any Hold* attributes, i.e. the lhs and rhs are evaluated before the comparison is performed.