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# Tough Difficult Tricky Operators in C Objective Questions and Answers

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 1 What is the output of the following program? ```#include void main() { int x=10,y=20; x=((x>9) && y=30 ? printf("\nTRUE"): printf("\nFALSE")); printf("\ny=%d",y); }``` A. FALSE y=20 B. TRUE y=30 C. Compilation Error D. None of the Above
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Explanation

Doubt: Are statements allowed in (?:)conditional operator ? Well, Yes Ofcourse! but without the semi-colon at the end!

Now, We might think that x>9 is TRUE and y=30 is TRUE Hence TRUE && TRUE becomes TRUE and TRUE is printed on screen, but we encounter

compile time error and the Error we get is :

error: lvalue required as left operand of assignment

lvalue: Any Variable that can be assigned a value and hence are permitted on the LHS of assignment operator.

rvalue: Any variable or constant whose value can be used in an expression or to assign value to lvalue.

It happens because the associativity of "and" operator "&&" is higher than Assignment Operator "=".

This converts the above statement in to

((x>9) && y) =30

i.e. TRUE && y = 30 // as y's current value is 20 , a non zero number,so

=> TRUE && TRUE =30

=> TRUE = 30 //  TRUE=1 by default so the expression becomes,

=> 1=30

And of course, A constant is not allowed on the Left hand side of the assignment operator as it is an rvalue!
To get correct output use parenthesis appropriately:
x=((x>9) && (y=30) ? printf("\nTRUE"): printf("\nFALSE"));

 2 ```int x=8,y=1; switch (x--,y++) { case 1: x *= 8; case 2:y *= x /= 2; case 3: case 4:y--; default: x += 5; } printf("%d %d", x, y);``` A. 64 2 B. 64 1 C. 33 55 D. 33 56
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Explanation

"," operator evaluates its left operand, evaluates its right operand and then returns the value of its right operand. So, (x--, y++) will evaluate x-- first then y++ and at last will return the value of y-- i.e. 1(post increment operator).

Latter steps are easy to calculate.

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# Operators in C Tough Difficult Tricky objective Questions and Answers

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