Operators

int i=10;

i=i++;//assigning a value is done after the expression is evaluated... here the expression value is "10 " since u r using post increment. after the expression evaluation is complete "i" value is 11. But as u have used the post increment the value of expression will be 10 which is assigned and printed.

System.out.println(i); Why 10?

And:

int i=10;

i=i++ + i++;

System.out.println(i); Why 21?

the expression here ((i++) + (i++)). We know that expression evaluation takes place from left to right.so at first (i++) it results in 10 and the second (i++) results to 11. so by adding up u will result in 21.

Jogeswara_Raoa at 2007-7-15 15:55:06 >

Think of it this way: if a ++ in front of/behind a field makes a copy of the field. The copy is the current field value without adding 1 if ++ is behind / with adding 1 if ++ is in front of the field.

i=i++is the same asi=i = 10

same with this example:

int i=10;

int j = 10;

i=i++ + j++;

i =i++ + j++is the same asi=i + j = 10+10 = 20

for the special case

int i=10;

i=i++ + i++;

you need to keep in mind that the first i++ copies the current value of i which is 10. The second i++ is also copying the current value of i but which is already 11 because of the first i++. Example:

i = i++ + i++

i = a + i++; //a is a copy of i that is 10, i itself is now 11

i = a + b; //b is a copy of i that is 11, i itself is now 12

i = 10 + 11 = 21

int i=10;

int j =i++ + i++;

System.out.println(i);

System.out.println(j);

MartinHilperta at 2007-7-15 15:55:06 >