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Question
UINT i,j;
i = j = 0;
i = ( i++ > ++j ) ? i++ : i--;

explain pls....
 Question Submitted By :: Boss
I also faced this Question!!     Rank Answer Posted By  
 
  Re: UINT i,j; i = j = 0; i = ( i++ > ++j ) ? i++ : i--; explain pls....
Answer
# 1
(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 ?    0 Yes 6 No
Chaneswara Reddy
 
  Re: UINT i,j; i = j = 0; i = ( i++ > ++j ) ? i++ : i--; explain pls....
Answer
# 2
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 ?    6 Yes 1 No
Raj
 
 
 
  Re: UINT i,j; i = j = 0; i = ( i++ > ++j ) ? i++ : i--; explain pls....
Answer
# 3
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 ?    4 Yes 0 No
Vignesh1988i
 

 
 
 
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