Sample Question

David Chan
Dear all,

I don't understand one sample question can anyone help me?

What will be print in the following code?

amethod() {
int i=0;
i = i++;
i = i++;
i = i++;
System.out.println(i);
}

The Answer is 0 will be printed.

Thanks.

David Chan

ur code is based on post-increment concept.
int i=0;
i=i++;
i=i++;
i=i++;
here variable i has been initialized to 0.In the 2nd statement,the value of i(ie. 0 )is assigned to i before i is incremented to 1.so now i will have 0 as its value.Again in the 3rd statement,the value of i(ie. 0) will be assigned to i before i is incremented to 1.so now i will have 0 and not 1 as its value.This process is continued for the 4th statement also.
Hence 0 gets printed.

if it had been the case of pre-increment like the below:
int i=0;
i=++i;
i=++i;
i=++i;
System.out.println(i);

then 3 would be the output since the value of i is incremented by 1 in all the 3 statements before being assinged to variable i again.

Hope this will help.
rajashree.

-Paul.

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After run the following code,

int i = 0;
int j = 0;

i = j++;

The result will be i = 0 and j = 1.

So, I think for the follow code, the result is the similar :

int i = 0;

i = i++;

i will be assigned with 0 then i increase by 1 then the result of i become 1.

I don't understand what different in the 2 situations.

Thanks

David Chan

With post-increment operators, Java evaluates the variable before the increment occurs. With the example i = i++, i on the right side of the assignment is evaluated as '0'; the post-increment value 0+1 is stored as the new value for 'i' then the assignment takes place i = 0 ... this effectively overwrites the previously stored value of '1'.

Hope that helps.

I'm going to move this over Certification Study as it is not related to Results.

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Jane Griscti
Sun Certified Programmer for the Java™ 2 Platform