Tag Archives: C Tutorials

Technical Aptitude Question – C Language – Part-2

1) main()
{
    clrscr();
}

clrscr();

Answer:
No output/error

Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

2) enum colors {BLACK,BLUE,GREEN}
main()
{
printf(“%d..%d..%d”,BLACK,BLUE,GREEN);
return(1);

}

Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

3) void main()
{
     char far *farther,*farthest;
     printf(“%d..%d”,sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer

4.) main()
{

int i=400,j=300;

printf(“%d..%d”);

}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf’s may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

5) main()
{
char *p;
p=”Hello”;
printf(“%c\n”,*&*p);
}

Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string “Hello”. *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

6) main()
{
        int i=1;
        while (i<=5)

         {
                  printf(“%d”,i); 

                  if(i>2)

                        goto here;
                   i++;
         }
}
fun()
{
here:
printf(“PP”);
}
Answer:
Compiler error: Undefined label ‘here’ in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label ‘here’ is available in function fun() Hence it is not visible in function main.

7) main()
{
       static char names[5][20]={“pascal”,”ada”,”cobol”,”fortran”,”perl”};
       int i;
      char *t;
       t=names[3];
      names[3]=names[4]; names[4]=t;
             for (i=0;i<=4;i++)
                   printf(“%s”,names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

8) void main()
{
      int i=5;
      printf(“%d”,i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i

9) void main()
{
      int i=5;
     printf(“%d”,i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

10) #include<stdio.h>
main()
{
     int i=1,j=2;

      switch(i)
     {
            case 1: printf(“GOOD”);
           break;

           case j: printf(“BAD”);
           break;
      }
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that
we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.

11) main()
{
      int i;
       printf(“%d”,scanf(“%d”,&i)); // value 10 is given as input here
}

Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

12) main ( )
{
     static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
     char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
      p = ptr;
     **++p;
      printf(“%s”,*–*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression
is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

13) main()
{
     int i=0;
     for(;i++;printf(“%d”,i)) ;
         printf(“%d”,i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is “evaluated”. Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

14) #include<stdio.h>
main()
{
     char s[]={‘a’,’b’,’c’,’\n’,’c’,”};
     char *p,*str,*str1;
     p=&s[3];
    str=p;
    str1=s;
    printf(“%d”,++*p + ++*str1-32);
}

Answer:
M
Explanation:
p is pointing to character ‘\n’.str1 is pointing to character ‘a’ ++*p meAnswer:”p is pointing to ‘\n’ and that is incremented by one.” the ASCII value of ‘\n’ is 10. then it is incremented to 11. the value of ++*p is 11. + +*str1 meAnswer:”str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77(“M”);

15) #include main()
{
      struct xx
     {
              int x=3;
             char name[]=”hello”;
      };
          struct xx *s=malloc(sizeof(struct xx));
          printf(“%d”,s->x);
          printf(“%s”,s->name);
 }
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

16) #include<stdio.h>
main()
{
       struct xx
         {
               int x;
              struct yy
              {
                       char s;
                       struct xx *p;
               };
              struct yy *q;
          };
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

17)  main()
{
        extern int i;
        i=20;
       printf(“%d”,sizeof(i));
}
Answer:
Linker error: undefined symbol ‘_i’.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn’t find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

18) main()
{
      printf(“%d”, out);
}

int out=100;

Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

19) main()
{
      extern out;
      printf(“%d”, out);
}

int out=100;
Answer:
100
Explanation:
This is the correct way of writing the previous program.

20) main()
{
     show();
}
void show()
{
     printf(“I’m the greatest”);
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn’t know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

21) main( )
{
      int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
     printf(“%u %u %u %d \n”,a,*a,**a,***a);
     printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.

2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output. for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a
+1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

22) main( )
{
     int a[ ] = {10,20,30,40,50},j,*p;
      for(j=0; j<5; j++)
     {
             printf(“%d” ,*a);
            a++;
     }

     p = a;
     for(j=0; j<5; j++)
    {
         printf(“%d ” ,*p);
         p++;
    }
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

23) main( )
{
      static int a[ ] = {0,1,2,3,4};
      int *p[ ] = {a,a+1,a+2,a+3,a+4};
      int **ptr = p;
      ptr++;
      printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
     *ptr++;
     printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
     *++ptr;
     printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
      ++*ptr;
      printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111

222

333

344
Explanation:

Let us consider the array and the two pointers with some address

a

0 1 2 3 4
100 102 104 106 108

p

100 102 104 106 108
1000 1002 1004 1006 1008

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is 1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

24) main( )
{
     char *q;
     int j;
      for (j=0; j<3; j++)
         scanf(“%s” ,(q+j));
       for (j=0; j<3; j++)
             printf(“%c” ,*(q+j));
      for (j=0; j<3; j++)
            printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M O U S E ‘\O’

When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M T R A C K ‘\O’

The third input starts filling from the location 102

M T V I R T U A L ‘\O’

This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

25) main( )
{
       void *vp;
      char ch = ‘g’,
     *cp = “goofy”;
      int j = 20;
      vp = &ch;
       printf(“%c”, *(char *)vp);
       vp = &j;
       printf(“%d”,*(int *)vp);
       vp = cp; printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C bits Manipulation

Bit manipulation is the act of algorithmically manipulating bits or other pieces of data shorter than a byte. C language is very efficient in manipulating bits.

Here are following operators to perform bits manipulation:

Bitwise Operators:

Bitwise operator works on bits and perform bit by bit operation.

Assume if B = 60; and B = 13; Now in binary format they will be as follows:

A = 0011 1100

B = 0000 1101

—————–

A&B = 0000 1000

A|B = 0011 1101

A^B = 0011 0001

~A  = 1100 0011

Try following example to understand all the Bitwise operators. Copy and paste following C program in test.c file and compile and run this program.
main()
{

   unsigned int a = 60;	/* 60 = 0011 1100 */  
   unsigned int b = 13;	/* 13 = 0000 1101 */
   int c = 0;           

   c = a & b;       /* 12 = 0000 1100 */ 
   printf("Line 1 - Value of c is %d\n", c );

   c = a | b;       /* 61 = 0011 1101 */
   printf("Line 2 - Value of c is %d\n", c );

   c = a ^ b;       /* 49 = 0011 0001 */
   printf("Line 3 - Value of c is %d\n", c );

   c = ~a;          /*-61 = 1100 0011 */
   printf("Line 4 - Value of c is %d\n", c );

   c = a << 2;     /* 240 = 1111 0000 */
   printf("Line 5 - Value of c is %d\n", c );

   c = a >> 2;     /* 15 = 0000 1111 */
   printf("Line 6 - Value of c is %d\n", c );
}

This will produce following result:-

Line 1 - Value of c is 12
Line 2 - Value of c is 61
Line 3 - Value of c is 49
Line 4 - Value of c is -61
Line 5 - Value of c is 240
Line 6 - Value of c is 15

There are following Bitwise operators supported by C language :-

 

Operator Description Example
& Binary AND Operator copies a bit to the result if it exists in both operands. (A & B) will give 12 which is 0000 1100
| Binary OR Operator copies a bit if it exists in eather operand. (A | B) will give 61 which is 0011 1101
^ Binary XOR Operator copies the bit if it is set in one operand but not both. (A ^ B) will give 49 which is 0011 0001
~ Binary Ones Complement Operator is unary and has the efect of ‘flipping’ bits. (~A ) will give -60 which is 1100 0011
<< Binary Left Shift Operator. The left operands value is moved left by the number of bits specified by the right operand. A << 2 will give 240 which is 1111 0000
>> Binary Right Shift Operator. The left operands value is moved right by the number of bits specified by the right operand. A >> 2 will give 15 which is 0000 1111

The shift operators perform appropriate shift by operator on the right to the operator on the left. The right operator must be positive. The vacated bits are filled with zero.

For example: x << 2 shifts the bits in x by 2 places to the left.

if x = 00000010 (binary) or 2 (decimal)

then: 
x >>= 2 => x = 00000000 or just 0 (decimal)

Also: if x = 00000010 (binary) or 2 (decimal) 
then
x <<= 2 => x = 00001000 or 8 (decimal)

 

Therefore a shift left is equivalent to a multiplication by 2. Similarly a shift right is equal to division by 2. Shifting is much faster than actual multiplication (*) or division (/) by 2. So if you want fast multiplications or division by 2 use shifts.

To illustrate many points of bitwise operators let us write a function, Bitcount, that counts bits set to 1 in an 8 bit number (unsigned char) passed as an argument to the function.

int bitcount(unsigned char x) 
{ 
   int count;

   for ( count=0; x != 0; x>>=1);
   {
      if ( x & 01)
         count++;
   }

   return count;
}

 

This function illustrates many C program points:

  • for loop not used for simple counting operation.
  • x >>= 1 => x = x>> 1;
  • for loop will repeatedly shift right x until x becomes 0
  • use expression evaluation of x & 01 to control if
  • x & 01 masks of 1st bit of x if this is 1 then count++

Bit Fields

Bit Fields allow the packing of data in a structure. This is especially useful when memory or data storage is at a premium. Typical examples:

  • Packing several objects into a machine word. e.g. 1 bit flags can be compacted.
  • Reading external file formats — non-standard file formats could be read in. E.g. 9 bit integers.

C allows us do this in a structure definition by putting :bit length after the variable.For example:

 

struct packed_struct {
  unsigned int f1:1;
  unsigned int f2:1;
  unsigned int f3:1;
  unsigned int f4:1;
  unsigned int type:4;
  unsigned int my_int:9;
} pack;

Here the packed_struct contains 6 members: Four 1 bit flags f1..f3, a 4 bit type and a 9 bit my_int.

C automatically packs the above bit fields as compactly as possible, provided that the maximum length of the field is less than or equal to the integer word length of the computer. If this is not the case then some compilers may allow memory overlap for the fields whilst other would store the next field in the next word.

 

 

Skip to toolbar