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

 
 

Option: B

Explanation :

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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

 
 

Option: C

Explanation :

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8:  

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

A.

1.8

B.

1.0

C.

2.0

D.

0.00

 
 

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;
}

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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

 
 

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.

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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 ));

outputs an integer that is same as (log in the answers are to the base two)

A.

an unpredictable value

B.

8 * log( x + 3 )

C.

log( x+ 3 )

D.

none of above

 
 

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)}.

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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

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