#c#ternaryoperator#function#operator

Question

Find the output ofthe following c code .

int fun1(int x);
int fun2(int x);

int main()
{
static int x=4;
x = (5>8?10:1!=2<5?fun1(++x): fun2(++x));
printf("Value of x:%d\n", x);
return 0;
}

int fun1(int x)
{
printf("x=%d\n", x);
return x;
}

int fun2(int x)
{
printf("x=%d\n", x);
return ++x;
}

(a) x=5
Value of x:6

(b) x=6
Value of x:6

(c) x=5

(d) None of these.
Value of x:5

Comments

I think ques is having some typo @Shubham Sharma 2 it should be 1!=2<5 , 11=2<5 doesn't mean anything. 

---

yes yes thank you..i am editing the question

Answer 1

Line: x = (5>8?10:1!=2<5?fun1(++x): fun2(++x)); consist multiple operators =, >,< ?:, !=. Hence, precednce and associativity of the operators needs to be identified to evaluate the expression.

Operator	Priority	Associativity
>, <	1	Left to Right
!=	2	Left to Right
?:	3	Right to Left
=	4	Right to Left

 

$x = \color{red}({\underbrace{\color{black}{\underbrace{5>8}}?10:\color{green}{\underbrace{\color{green}{\underbrace{1!= \color{green}{\underbrace{2<5}}}}?\color{blue}{fun1(++x)}\color{green}{:}\color{magenta}{ fun2(++x)}}_{\text{due to Right associativity of Ternary Operator}}}}});$

 

Order of evaluation of ternary operator 

Source :K&R pg no. 51

 

The following ternary operation can be viewed like this :

if(5>8){
    x=10;
}
else if(1!=2<5){
    fun1(++x);
}
else{
    fun2(++x);
}

Since (5<8) so the else if resolves , now the following condition is evaluated as (1!=(2<5)) because $<$  has higher precedence than $!=$ . (1!=(2<5)) will give result as 0 as 1==1 and now this condition becomes false and we will resolve the else statement which will call the fun2 with increment value of x i.e $5$ .

$ fun2 $ will print the output $x=5$ and will return the updated value 6 , now  $var x$ of main will get the values as 6 , and it will print the output as $value of x =6$.

Answer: A

Comments

Precedence of <, > and ternary operator

Associativity of ternary operator are key points to evaluate the expression.

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