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# C Programming MCQ - I/O Operations

6:

printf( "%c", 100);

 A. prints 100 B. prints ASCII equivalent of 100 C. prints garbage D. none of above Answer Report Discuss Option: B Explanation : Click on Discuss to view users comments. Write your comments here:
7:

The program fragment

int i = 263 ;
putchar( i );

 A. prints 263 B. prints the ASCll equivalent of 263 C. rings the bell D. prints garbage Answer Report Discuss Option: C Explanation : Click on Discuss to view users comments. Write your comments here:
8:

The following statement
prlntf( ‘%f ’, 9/5) ;
prints

 A. 1.8 B. 1.0 C. 2.0 D. 0.00 Answer Report Discuss Option: D Explanation : This statement is undefined behavior because the argument has to be of type double and 9 / 5 is of type int. Following program yields an output of 1.8 int main() {     printf("%f",9/5.0);  return 0; } Click on Discuss to view users comments. Write your comments here:
9:

The following program fragment

unsigned i = 1;
int j = -4;
printf ("%u ", i + j);

prints

 A. garbage B. -3 C. an integer that changes from machine to machine D. none of above Answer Report Discuss Option: C Explanation : In the computer I used to execute this program. the output was 4294967293. That's because in my system, sizeof (int ) is 4 bytes (32 bits), and negative numbers are represented in 2's complement form. This means -4 will be represented as 11111111 11111111 11111111 11111100 (i.e. 30 one's followed by 2 zeroes). Note that this number is 232 - 1 - 3. Before j gets added to 1, it will be converted to an unsigned integer. So, i + j is essentially adding 1 to 232- 1 - 3 which gives 4294967293. Click on Discuss to view users comments. Write your comments here:
10:

If the following program fragment ( assume negative numbers are stored in 2's complement form )
unsigned i=1;
int j = -4;
printf( " %u ", i + j);
prints x, then printf( " %d ", 8* sizeof( int ));

 A. an unpredictable value B. 8 * log( x + 3 ) C. log( x+ 3 ) D. none of above Answer Report Discuss Option: C Explanation : Let size of ( int ) = 1. So, -4 will be stored as 1 1 1 1 1 1 0 0. Since we are adding unsigned and signed integers, the signed gets converted to unsigned. So. i + j will become 1 1 1 1 1 1 0 1. We are trying to print this as an unsigned integer. So, what is printed will be 28 - 1 - 2. So, log (x + 3) = 8 (i.e.. 8 * sizeof ( int)}. Click on Discuss to view users comments. Write your comments here:

Syllabus Covered in this section is

• Elements of C-Tokens
• Identifiers
• data types in C
• Control structures in, C
•  Sequence, selection and iteration(s).
• Structured data types in C
• Arrays, Structures, union, String, and pointers
• Functions, Recursion
• Parameter passing, Scope
•  Binding, Abstract data types

This Section covers Multiple Choice Questions Answers in C Programming . Here questions answers are given with explanation and references. These questions can be used for the preparation of various competitive and academic exams.

Who can benefit -

• Any undergraduate or postgraduate student who is seeking C Programming objective type questions answers can use this section.
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• C Programming MCQs can also be used by the students who are pursuing B.Sc or Msc Computer Science.
• C Programming Questions Answers can also be used by BCA students for the preparation of their exams.
• Any student who is pursuing B.Sc. in Information Technology can also use this C Programming mcq Questions Answers.
•  MCA students can also prepare for their exams using  C Programming Objective Type Questions Answers.

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