Languages-C & C Plus Plus [3i Infotech Placement]: Sample Questions 264 - 265 of 354

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Question 264

C & C Plus Plus
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Describe in Detail

Essay▾

What is the output of the following program?

1. main ()
2. {
3.     inti =4, j =7;
4.     j =j|| i ++ && printf ( “YOUCAN” );
5.     printf ( “%d %d”, i, j);
6. }

Explanation

In the program

 int i = 4, j = 7; Given the integer variable i = 4 and j = 7 j = j|| i ++ && printf ( “YOU CAN” ) ; The boolean expression needs to be evaluated only till the truth value of the expression is not known.J is not equal to zero itself means that the expression՚s truth value is 1.Because it is followed by || and true || (anything) ⩾ true, thus (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of I remains the same.Similarly when && operator is involved in an expression, it becomes false any of the operands become false- the whole expression՚s truth value becomes false and hence the remaining expression is notevaluated. Thus, false && (anything) ⩾ false where (anything) will not be evaluated. printf ( “% d % d” , i, j) ; printf prints the value of i = 4 and j = 1

Question 265

C & C Plus Plus
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Write in Short

What is the output of the following program?

1. main ()
2. {
3.     unsignedinti =10;
4.     while (i-->= 0)
5.     printf ( “%u” i);
6. }

Explanation

10 9 8 7 6 5 4 3 2 1 0 65535 65534 …

In the program

 unsigned int i = 10; Since i is an unsigned integer it can never become negative. while (i-- >= 0)printf ( “% u” i) ; So the expression i-- >= 0 will always be true, leading to an infinite loop.

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