Consider
int b = 2; int[] a = new int[4]; a[a[b]] = a[b] = b = 2; for (int i = 0; i <= 3; i++) { System.out.println(a[i]); } The output is
2 0 2 0 I was expecting a[0] to be zero.
- 4Why would you ever have such code? Any statement with three assignments in is a bad idea...Jon Skeet– Jon Skeet2014-03-25 10:46:36 +00:00Commented Mar 25, 2014 at 10:46
- @Pshemo: Great spot and the accepted answer on the duplicate is excellent.Bathsheba– Bathsheba2014-03-25 11:42:00 +00:00Commented Mar 25, 2014 at 11:42
- @Bathsheba All credits to TheEnd. He found it and posted link in his answer. I just flagged it as duplicate.Pshemo– Pshemo2014-03-25 11:44:39 +00:00Commented Mar 25, 2014 at 11:44
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6 Answers
Excerpt from JLS 15.26.1. Simple Assignment Operator =
If the left-hand operand is an array access expression (§15.13), possibly enclosed in one or more pairs of parentheses, then:
First, the array reference subexpression of the left-hand operand array access expression is evaluated. If this evaluation completes abruptly, then the assignment expression completes abruptly for the same reason; the index subexpression (of the left-hand operand array access expression) and the right-hand operand are not evaluated and no assignment occurs.
This means that a[a[b]] evaluates to a[0] as it's evaluated first. Then we proceed simply as a[0] = a[b] = b = 2, with the assignment taking place from right to left.
See http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.26.1
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Because a[a[b]] is same as a[0] and you are assigning value 2 for that..
b = 2, a[b] = 0 so a[a[b]] = 2 answered Mar 25, 2014 at 10:47
Abimaran Kugathasan
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a[a[b]] = a[b]=b=2 , This executes from left hand side
First
a[a[b]] i.e, a[a[2]] -> a[0] //initially the value of a[2] will be 0 now,
a[0]=a[2]=b=2; so the output is,
2 0 2 0 
Comments
I guess this thread might help you to understand the evaluation order in java: What are the rules for evaluation order in Java?
The most important part of this problem is that the first = is the actual assignment. In java the index is always evaluated befor the asignment, therefor a[a[b]] is the fist one in the evaluation order. And at this point a[b] is 0.
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Let's see what a[a[b]] = a[b]=b=2; does in detail:
a[a[b]] : b is 2, a[2] is 0, a[a[2]] is a[0].
so a[a[b]] = a[b]=b=2; is evaluated to a[0] = a[b]=b=2;
that is evaluated to a[0] = a[2] =2;, so a[2] is 2 and a[0] is equal to a[2]
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I think you are reading
a[a[b]] = a[b]=b=2;
from right to left and expecting the array and b to be reevaluated at each step, when in fact it seems like a and b are frozen at the time of the assignment. Consider the equivalent code:
1) int b = 2; //array initialized to 0 2) int[] a= new int[4]; // 0 0 0 0 // at this point a[b] is a[2] = 0, therefore below a[a[b]]=2 is equivalent to a[0]=2 3) a[a[b]] = 2 // 2 0 0 0 // the same: a[b] = 2 is equivalent to a[2] = 2 -note that the arrays are zero based 4) a[b]= 2 // 2 0 2 0 5) b=2; 