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'.

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

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.

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.

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.

errors are known as?

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