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how the assignment happened

 
Michael Lin
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public class Arg{
public static void main(String argv[]){
Arg inc = new Arg();
int i =0;
i = i++;
System.out.println(i); //line 1, 0???

int k =0;
System.out.println(k++); //0
System.out.println(k); //line3, 1
}
}
//hi, my confusion is why the line 1 still give 0?
//after i= i++, the i should be 1 now even if it is post increment.
//it should like the same in the line 3.
Someone please explain to me, thanks!
 
rajani peddi
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Michael Lin,
int i =0; // line 1
i = i++; // line 2

There are 3 steps involved in this :
1. First in line2 i is assigned 0 (cos of line 1, and i++ is a postincrement operator)
2. Then i++ is evaluated ( so i = 1)
3. Now i = 0 is evaluated ( from step 1)
Hope this clears
rajani
 
vadiraj vd
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Hi micheal,
I have he same doubt that how the assignment is happening.
I tried this which is the equivalent of your code.
public class StrangeAssignment
{
public static void main(String a[])
{
int i = 0;
i = i = i+1; // i++ is equivalent to i=i+1.
System.out.println(i);
}
}
But the answer is 1 !!. Really it is a strange assignment.
I tried the same in C/C++ it's giving 1.
Regards,
vadiraj

 
anil bisht
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ok lets take this expression
i = i++ + i
what is the answer .. its one..
what it did was
it writes the value of i before the + sign ( i=0) which is zero till now and then increments the value of i by 1
our expresion becoms
i = 0 +... and i becomes 1
so now the value of i after + sign is assigned 1( as now i = 1)
so the final expression becomes ( till now i is one)
i = 0 + 1
now calculation the expression we get
i = 1
if u got this now we will go back to ur question
now our expression is
i = i++
first i=0 so after the = sign i is assigned 0 and the value of i is incremented by 1 and i becomes 1
and our expression becomes
i = 0 ( till now the value of i is one)
but after evaluating this expression i becomes 0
HTH
anil
 
Sudhir Thorat
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Does this mean that the following statements
public class Postincrement{
public static void main(String args[]){
i=0;
i=i++ + i++;
System.out.println(i) ;
}
}
would print 0 but the value of the variable i is 2 ?
Thanks
Sudhir
 
anjan bhushan
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I think confusion is due to behaviour of operator in C.
In C it will do
i= 0;
i=i++;/* 2 */
printf("%d",i);
ans i =1
C do it as assign i=0 on line two then increment it by 1.
Java seems workingin different way
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