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please explain

 
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What gets printed when the following program is compiled and run. Select the one correct answer.

1.class test {
public static void main(String args[]) {
int i,j,k,l=0;
k = l++;
j = ++k;
i = j++;
System.out.println(i);
}
}


0
1
2
answer is 1.

2.public class test {
public static void main(String args[]) {
int i, j;
int k = 0;
j = 2;
k = j = i = 1;
System.out.println(k);
}
}


The program does not compile as k is being read without being initialized.
The program does not compile because of the statement k = j = i = 1;
The program compiles and runs printing 0.
The program compiles and runs printing 1.
The program compiles and runs printing 2.

answer is it prints 2.
 
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Hi Ketki,
++ operator changes the outcome of an expression depending on whether the operator is placed before or after the operand. At line, where k=l++ the value of l is first assigned to k and then the value of l is incremented so k=0. And in the next line the value of k is incremented and then it is assigned to j so j=1. At last the value of j is assigned to i and j is incremented so i=1.


For the second question the output is one not two
 
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the answer for the first question is 1...
intially all the values are 0...
i=0, j =0, k=0, l=0
k=l++ >>>>> k = l which is 0(and then increment l by 1)
so k=0 and l =1

j=++k >>>>> j = (k+1) which is 1 (and remember k value is now 1)

i=j++ >>>>> i= j which is 1 ( and then increment j by 1)

so when you print i its value is 1



for the second question the answer is it prints 1....

because its intializing from right to left
[ June 12, 2007: Message edited by: suresh koutam ]
[ June 12, 2007: Message edited by: suresh koutam ]
 
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