What is the output

[Sneha Kapoor]

 

public class Inc{
public static void main(String argv[]){
                Inc inc = new Inc();
                int i =0; 
                inc.fermin(i);
                i = i++;
                System.out.println(i);
				System.out.println(i);
        }
        void fermin(int i){
                i++;
        }
}

why the i output as 0..when it will print as 1

[Joshua Ebarvia]

Hi, I'm not an expert here but I'll try to answer your question.

#
# public class Inc{
# public static void main(String argv[]){
# Inc inc = new Inc();
# int i =0;
# inc.fermin(i);
# i = i++;
# System.out.println(i);
# System.out.println(i);
# }
# void fermin(int i){
# i++;
# }
# }

When this is compiled and run, the output is

0
0

It is just as expected, though you have passed variable i from the main method to method fermin() and used the i++, the ++ operator just took effect inside the method fermin. After the method is done, the effect doesn't change your variable i in the main method. when you reach the code i = i++, the variable i still has the value of 0 since i++ leads to 0. Then when you print the i using the println, effect of the ++ was not considered since you have assigned it to i itself.

[Venkata Kumar]

Hi Sneha,

In java the parameters to method arguments are passed by value

 
  void fermin(int i){  
                 i++;  
   }  

The parameter i is local to method fermin and incrementing this value won't affect the variable i which is declared in main program.

Please see the kink http://www.coderanch.com/t/408774/Java-General-beginner/java/could-anyone-explain-this-please

[Ankit Garg]

Please when you post a question, then remember to Quote Your Sources...

[Sneha Kapoor]

the variable i still has the value of 0 since i++ leads to 0.

how come i = i++ leads to 0

[Joshua Ebarvia]

int i =0; // initial value of i in main()
inc.fermin(i); //passes i to fermin() but doesn't have an effect to i in main()
i = i++; // since i = 0 and you have the post increment operator ++, it is like saying increment i on the next operation, but not this time
System.out.println(i) // still leads to zero because of i = i++

[Sneha Kapoor]

thanks got it so if i do below things so i will get i = 1

i=i+1;
System.out.println(i);

[Joshua Ebarvia]

Sneha Kapoor wrote:thanks got it so if i do below things so i will get i = 1

i=i+1;
System.out.println(i);

yes...l. you can also try

   i = ++i //it is like saying add one to my present value. Since my value is zero, I'm now one.. :)

[Bob Wheeler]

Sneha Kapoor wrote:thanks got it so if i do below things so i will get i = 1

i=i+1;
System.out.println(i);

Or

 
// two independent lines
++i;   // next line i=1
i++ ;   // next line i=1

cheers
Bob

[aabir sanyal]

I am unable to get this, then what is the effect of i=i++;

Really confusing

[Bob Wheeler]

aabir sanyal wrote:I am unable to get this, then what is the effect of i=i++;

Really confusing

check this out

[aabir sanyal]

Bob Wheeler wrote:

aabir sanyal wrote:I am unable to get this, then what is the effect of i=i++;

Really confusing

check this out

Thanks Bob