What are the order of operations when using two assignment operators in a single line?
public static void main(String[] args){
int i = 0;
int[] a = {3, 6};
a[i] = i = 9; // this line in particular
System.out.println(i + " " + a[0] + " " + a[1]);
}
Edit: Thanks for the posts. I get that = takes values from the right, but when I compile this I get:
9 9 6
I thought it would have been and ArrayOutOfBounds exception, but it is assigning 'a[i]' before it's moving over the 9. Does it just do that for arrays?
= is parsed as right-associative, but order of evaluation is left-to-right.
ReplyDeleteSo: The statement is parsed as a[i] = (i = 9). However, the expression i in a[i] is evaluated before the right hand side (i = 9), when i is still 0.
It's the equivalent of something like:
int[] #0 = a;
int #1 = i;
int #2 = 9;
i = #2;
#0[#1] = #2;
As per the specs:
ReplyDeletehttp://docs.oracle.com/javase/specs/jls/se5.0/html/expressions.html
15.26 Assignment Operators
There are 12 assignment operators; all are syntactically right-associative (they group right-to-left). Thus, a=b=c means a=(b=c), which assigns the value of c to b and then assigns the value of b to a.
So, a[i] = i = 9; is the same as i = 9; a[i] = i;
If I remember correctly, = operator is right-associative; so i will be assigned first, then a[i].
ReplyDeleteThe = operator is right-associative(as others have already said). This can be easily proven with this test:
ReplyDeleteint i = 2;
int j = 3;
int x = i = j;
System.out.println(x); // This prints out 3.
This works with all types, Objects and primitives.
The way I have heard this referred to is "dual assignment", since, using the example above, you are assigning the value of j to both i and x.
What exactly do you espect to get as an output? There, you are assigning 9 to i and then the value of i to the array (but the array position was already calculed to the 0 position); so, since maybe you are confused about the array position in which the 9 is stored?... First you get a pointer to the a[0] (current value of i) and then, the compiler goes to process the rest of the sentence.
ReplyDelete