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PODouble structural equality operators: if(a==b==c)
 

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    <p>I wrote some code by accident today and was surprised when Eclipse did not yell at me, for once. The code had a double use of the structural equality operator (<code>==</code>) similar to the below with the <code>if(a==b==c)</code> structure.</p> <pre><code>public class tripleEqual { public static void main(String[] args) { boolean[] a = { true, false }; boolean[] b = { true, false }; boolean[] c = { true, false }; for (int aDex = 0; aDex &lt; 2; aDex++) { for (int bDex = 0; bDex &lt; 2; bDex++) { for (int cDex = 0; cDex &lt; 2; cDex++) { if (a[aDex] == b[bDex] == c[cDex]) { System.out.printf("Got a==b==c with %d %d %d\n", aDex, bDex, cDex); } } } } } } </code></pre> <p><strong>The output is</strong> </p> <pre><code>Got a==b==c with 0 0 0 Got a==b==c with 0 1 1 Got a==b==c with 1 0 1 Got a==b==c with 1 1 0 </code></pre> <p>Playing around, I notice I can't do <code>if(a==b==c)</code> with any type but <code>boolean</code>. From that the boolean expression is </p> <pre><code>( A'. B'. C') + ( A'. B . C ) + ( A . B'. C ) + ( A . B . C') </code></pre> <p>which simplifies to <code>(A=B).'C + (A&lt;&gt;B).C</code>. </p> <p>Thus, ignoring side-effect, <code>if(a==b==c)</code> is equal to <code>if(a==b &amp;&amp; !c) || (a!=b &amp;&amp; c))</code>. </p> <p>Can anyone explain how the <code>if(a==b==c)</code> syntax suggests that?</p> <p><strong>Edit 1:</strong></p> <p>I found where my confusion was after so many people explained the left-associativity. Usually I write '1' for true and '0' for false but my minimized truth table/output in the above test, I had '0' for true and '1' for false. The negation of the expression <code>( A'. B'. C') + ( A'. B . C ) + ( A . B'. C ) + ( A . B . C')</code> is <code>(A=B)=C</code>!</p>
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    1. COI've answered on a similar question a week ago (the question was about JavaScript, but the rules are the same): http://stackoverflow.com/questions/19955573/why-is-1-1-1-false/19955600#19955600
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    2. CO@John Those do evaluate to false, so I don't really follow what you are saying. I will say however my saying "same idea as 1+2+3" is misleading, as `+` over integers is associative and `==` over booleans is not (ie. ((a + b) + c) equals (a + (b + c)) over integers, but ((a == b) == c) does not equal (a == (b == c)) over booleans)
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    3. CO@bengoesboom I do see where my confusion was, in my test code I had the output being 0=true & 1=false. Which is the opposite of what I am used to. Negating (A=B).'C + (A<>B).C does give me (a=b)=c.
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