Languages-C & C Plus Plus [3i Infotech Placement]: Sample Questions 143 - 144 of 354

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Question 143

C & C Plus Plus
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Write in Short

What is the output of the following program?

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

Explanation

In the program

 int i = 5, j = 6, z; Given the integer variables i = 5, j = 6 and z printf ( “% d” i ++ + j) ; Expression i ++ + j is treated as (i ++ + j)So (5 + 6 = 11)
• I ++ + j is equivalent to i ++ + j.
• Not because of operator precedence but because of grammar rules, which are generally greedy.
• The ++ + is resolved to ++ + by the compiler before expressions are parsed.
• So the answer is 11

Question 144

C & C Plus Plus
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Describe in Detail

Essay▾
1. voidmain ()
2. {
3.     unsignedi =1;
4.     /⚹unsigned char k =-1 ➾ k =255; ⚹/
5.     signedj =-1;
6.     /⚹char k =-1 ➾ k =65535 ⚹/
7.     /⚹unsigned or signed int k =-1 ➾ k =65535 ⚹/
8.     if (i<j)
9.     printf("less");
10.     elseif (i>j)
11.     printf ("greater");
12.     elseif (i==j)
13.     printf ("equal");
14. }

Explanation

• Program defines signed and unsigned values
 unsigned i = 1; define the unsigned variable i = 1 signed j =-1; define the signed variable j =-1 if (i < j)printf ( “less” ) ;else if (i > j)printf ( “greater” ) ;else if (i == j)printf ( “equal” ) ;} check the if conditionif (1 <-1)printf ( “less” ) ;printf prints “less” because unsigned can hold a larger positive value, and no negative value.Signed integers can hold both positive and negative numbers.Unsigned uses the leading bit as a part of the value, while the signed version uses the left-most-bit to identify if the number is positive or negative.

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