Possible Duplicate:
Java - Order of Operations - Using Two Assignment Operators in a Single Line
If we assign a variable a value in a chain like the following,
int x=10, y=15;
int z=x=y;
System.out.println(x+" : "+y+" : "+z);
then the value of all three variables x
, y
and z
becomes 15
.
I however don't understand the following phenomenon with an array.
int array[]={10, 20, 30, 40, 50};
int i = 4;
array[i] = i = 0;
System.out.println(array[0]+" : "+array[1]+" : "+array[2]+" : "+array[3]+" : "+array[4]);
It outputs 10 : 20 : 30 : 40 : 0
. It replaces the value of the last element which is array[4]
with 0
.
Regarding previous assignment statement - int z=x=y;
, I expect the value of the first element means array[0]
to be replaced with 0
. Why is it not the case? It's simple but I can't figure it out. Could you please explain?
By the way, this assignment statement array[i] = i = 0;
is dummy and it has no value of its own in this code and should no longer be used but I just wanted to know how thing actually works in this case.
int i = 4;
when i equals to 4 the array[i]
equals to array[4]
so array[i] = i = 0;
is equivalent to array[4] = i = 0;
. That is way it change the value of index 4 with 0.
The separators [] and () change precedence. Everything within these separators will be computed before Java looks outside them.
array[i] = i = 0;
During compiler phases, the first change to this line will happen as follows:
array[4] = i = 0; // subscript separator is evaluated first.
Now, assignment operation is right-associative, So, i
is assigned value 0
and then array[4] is assigned value of i
i.e. 0
.
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