UGC NET CSE | June 2016 | Part 2 | Question: 11

Question

Given i=0, j=1, k=-1, x=0.5, y=0.0, what is the output of the following expression in C language?

x*y<i+j $\mid \mid$k

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer : 1

x*y < i + j || k 

if we go a/c to precedence and associativity then lets parenthesize it 

(((x*y) <( i + j)) || k) 

* Has higher precedence then + and a/c to associativity they are left to right 

comparison operator has less precedence then "*" and "+" .

Logical Or has lest precedence among all the operator .So

first we calculate (x*y) then ( i + j) after that ((x*y) <( i + j)) and finally

(((x*y) <( i + j)) || k) 

(x*y) = 0       ( i + j) = 1 

(( 0.0 < 1 ) || k ) Here LHS holds true  means we don't care what is the value of k it will always be True in case of OR .So Answer must be 1.

Answer 2

x+y<i+j || k

0.5(0) consider it as interger + 0.0 < 1 + 0 is true  

so output is 1