UINT i,j;
i = j = 0;
i = ( i++ > ++j ) ? i++ : i--;
explain pls....
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Answer / raj
1.we know that i=j=0 initially
2.then it will checks the non-incremented 'i' value(i.e 0)
with incremented 'j' value(i.e 1). So obviously condition
is falls.
3.now false statement has to be executed i.e (i--),before
executing this 'i' value is (i.e incremented value)'1'
after executing false condition(i.e i--)the value of 'i'
becomes '0'.
4.So the value of 'i' is '0'.
| Is This Answer Correct ? | 30 Yes | 2 No |
Answer / vignesh1988i
UNIT i,j :
this line indicates that UNIT is an user defined data type. it may been declared as follows :
typedf int UNIT
we are making the code more readable
i=j=0 : indicates that the var. i and j are declared as 0
i=(i++>++j) ? i++ : i-- : the process here is
i++ is an post incrementation . if this is compared with any relational or any operaters first that value will be operated first and the 'i' will get incremented ........
but ++j if we take first thing it will increment the value and then operation will be performed
so when it is compared first i will be 0 and j will be 1 so 0 is not greater than 1. so false, so it will go to the statement after ':' so i-- is there so final value of i will be 0.
thank u
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / dhatchina moorthy
r u guys nuts the person answered first is right.he
non-incremented 'i' value(i.e 0)
with incremented 'j' value(i.e 1). So condition
is falls.
so false part has to be executed i.e (i--),before
executing this 'i' value is (i.e incremented value)'1'
after executing false condition(i.e i-- her it doesn't
increment bcoz it is postfix operator)the value of 'i'
becomes '1'.
still having doubts, compile this program in ur pc.
| Is This Answer Correct ? | 5 Yes | 1 No |
Answer / prem_mallappa
All answers are incorrect except this one.
Read c-faqs (Frequently asked questions about C)
The answer is unpredictable or implementation defined.
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / chaneswara reddy
(i++ > j++) gives 0 because 0 > 0 is false so it return 0.
before returning 0 i is 1 ,but it is overwrite by 0.
In the Conditional operator false means ,it executes i++;
so i is 1.
| Is This Answer Correct ? | 3 Yes | 19 No |
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