Friday, 17 October 2014

C questions answers

.
 What will be output if you will compile and execute the following c code?


void main(){
   int i=320;
   char *ptr=(char *)&i;
   printf("%d",*ptr);
}

(A) 320
(B) 1
(C) 64
(D) Compiler error
(E) None of above


Explanation:
As we know size of int data type is two byte while char pointer can pointer one byte at time.
Memory representation of int i=320

So char pointer ptr is pointing to only first byte as shown above figure.
*ptr i.e. content of first byte is 01000000 and its decimal value is 64.
Data type tutorial. 

2.

What will be output if you will compile and execute the following c code?

#define x 5+2
void main(){
    int i;
    i=x*x*x;
    printf("%d",i);
}

(A) 343
(B) 27
(C) 133
(D) Compiler error
(E) None of above


Explanation:
As we know #define is token pasting preprocessor it only paste the value of micro constant in the program before the actual compilation start. If you will see intermediate file you will find:
test.c 1:
test.c 2: void main(){
test.c 3: int i;
test.c 4: i=5+2*5+2*5+2;
test.c 5: printf("%d",i);
test.c 6: }
test.c 7:
You can absorb #define only pastes the 5+2 in place of x in program. So,
i=5+2*5+2*5+2
=5+10+10+2
=27
Preprocessor tutorial.

3.

What will be output if you will compile and execute the following c code?

void main(){
char c=125;
    c=c+10;
    printf("%d",c);
}

(A) 135
(B) +INF
(C) -121
(D) -8
(E) Compiler error

Explanation:
As we know char data type shows cyclic properties i.e. if you will increase or decrease the char variables beyond its maximum or minimum value respectively it will repeat same value according to following cyclic order:

So,
125+1= 126
125+2= 127
125+3=-128
125+4=-127
125+5=-126
125+6=-125
125+7=-124
125+8=-123
125+9=-122
125+10=-121

4.

What will be output if you will compile and execute the following c code?

void main(){
   float a=5.2;
  if(a==5.2)
     printf("Equal");
  else if(a<5.2)
     printf("Less than");
  else
     printf("Greater than");
}

(A) Equal
(B) Less than
(C) Greater than
(D) Compiler error
(E) None of above


Explanation:
5.2 is double constant in c. In c size of double data is 8 byte while a is float variable. Size of float variable is 4 byte.
So double constant 5.2 is stored in memory as:
101.00 11001100 11001100 11001100 11001100 11001100 11001101
Content of variable a will store in the memory as:
101.00110 01100110 01100110
It is clear variable a is less than double constant 5.2
Since 5.2 is recurring float number so it different for float and double. Number likes 4.5, 3.25, 5.0 will store same values in float and double data type.
Note: In memory float and double data is stored in completely different way. If you want to see actual memory representation goes to question number (60) and (61).
Data type tutorial.

5.

What will be output if you will compile and execute the following c code?

void main(){
  int i=4,x;
  x=++i + ++i + ++i;
  printf("%d",x);
}

(A) 21
(B) 18
(C) 12
(D) Compiler error
(E) None of above


Explanation:
In ++a, ++ is pre increment operator. In any mathematical expression pre increment operator first increment the variable up to break point then starts assigning the final value to all variable.
Step 1: Increment the variable I up to break point.

Step 2: Start assigning final value 7 to all variable i in the expression.


So, i=7+7+7=21
Operator tutorial.

6.

What will be output if you will compile and execute the following c code?

void main(){
 int a=2;
 if(a==2){
   a=~a+2<<1;
   printf("%d",a);
 }
 else{
  break;
 }
}

(A) It will print nothing.
(B) -3
(C) -2
(D) 1
(E) Compiler error


Explanation:
Keyword break is not part of if-else statement. Hence it will show compiler error: Misplaced break
Control statement tutorial

7.

What will be output if you will compile and execute the following c code?

void main(){
  int a=10;
  printf("%d %d %d",a,a++,++a);
}

(A) 12 11 11
(B) 12 10 10
(C) 11 11 12
(D) 10 10 12
(E) Compiler error


Explanation:
In c printf function follows cdecl parameter passing scheme. In this scheme parameter is passed from right to left direction.

So first ++a will pass and value of variable will be a=10 then a++ will pass now value variable will be a=10 and at the end a will pass and value of a will be a=12.
Function tutorial.

8.

What will be output if you will compile and execute the following c code?

void main(){
   char *str="Hello world";
   printf("%d",printf("%s",str));
}

(A) 11Hello world
(B) 10Hello world
(C) Hello world10
(D) Hello world11
(E) Compiler error


Explanation:
Return type of printf function is integer and value of this integer is exactly equal to number of character including white space printf function prints. So, printf(“Hello world”) will return 13.

9.

What will be output if you will compile and execute the following c code?

#include "stdio.h"
#include "string.h"
void main(){
   char *str=NULL;
   strcpy(str,"cquestionbank");
   printf("%s",str);
}

(A) cquestionbank
(B) cquestionbank\0
(C) (null)
(D) It will print nothing
(E) Compiler error


Explanation:
We cannot copy any thing using strcpy function to the character pointer pointing to NULL.
More questions of string.

10.

What will be output if you will compile and execute the following c code?

#include "stdio.h"
#include "string.h"
void main(){
  int i=0;
  for(;i<=2;)
   printf(" %d",++i);
}

(A) 0 1 2
(B) 0 1 2 3
(C) 1 2 3
(D) Compiler error
(E) Infinite loop


Explanation:
In for loop each part is optional.
Complete tutorial of looping in C.

11.

What will be output if you will compile and execute the following c code?

void main(){
  int x;
  for(x=1;x<=5;x++);
    printf("%d",x);
}

(A) 4
(B) 5
(C) 6
(D) Compiler error
(E) None of above


Explanation:
Body of for loop is optional. In this question for loop will execute until value of variable x became six and condition became false.
Looping tutorial.

12.

What will be output if you will compile and execute the following c code?

void main(){
printf("%d",sizeof(5.2));
}

(A) 2
(B) 4
(C) 8
(D) 10
(E) Compiler error


Explanation:
Default type of floating point constant is double. So 5.2 is double constant and its size is 8 byte.
Detail explanation of all types of constant in C.

13.

What will be output if you will compile and execute the following c code?

#include "stdio.h"
#include "string.h"
void main(){
char c='\08';
printf("%d",c);
}


(A) 8
(B) ’8’
(C) 9
(D) null
(E) Compiler error


Explanation:
In c any character is starting with character ‘\’ represents octal number in character. As we know octal digits are: 0, 1, 2, 3, 4, 5, 6, and 7. So 8 is not an octal digit. Hence ‘\08’ is invalid octal character constant.
Hexadecimal character constant.

14.

What will be output if you will compile and execute the following c code?

#define call(x,y) x##y
void main(){
int x=5,y=10,xy=20;
printf("%d",xy+call(x,y));
}

(A) 35
(B) 510
(C) 15
(D) 40
(E) None of above


Explanation:
## is concatenation c preprocessor operator. It only concatenates the operands i.e.
a##b=ab
If you will see intermediate file then you will find code has converted into following intermediate code before the start of actual compilation.
Intermediate file:
test.c 1:
test.c 2: void main(){
test.c 3: int x=5,y=10,xy=20;
test.c 4: printf("%d",xy+xy);
test.c 5: }
test.c 6:
It is clear call(x, y) has replaced by xy.
Preprocessor tutorial.

15.

What will be output if you will compile and execute the following c code?


int * call();
void main(){
int *ptr;
ptr=call();
clrscr();
printf("%d",*ptr);
}
int * call(){
int a=25;
a++;
return &a;
}


(A) 25
(B) 26
(C) Any address
(D) Garbage value
(E) Compiler error


Explanation:
In this question variable a is a local variable and its scope and visibility is within the function call. After returning the address of a by function call variable a became dead while pointer ptr is still pointing to address of variable a. This problem is known as dangling pointer problem.
Complete pointer tutorial.

16.

What is error in following declaration?

struct outer{
int a;
struct inner{
char c;
};
};

(A) Nesting of structure is not allowed in c.
(B)
It is necessary to initialize the member variable.
(C) Inner structure must have name.
(D) Outer structure must have name.
(E) There is not any error.


Explanation:
It is necessary to assign name of inner structure at the time of declaration other wise we cannot access the member of inner structure. So correct declaration is:
struct outer{
int a;
struct inner{
char c;
}name;
};
Union tutorial.

17.

What will be output if you will compile and execute the following c code?

void main(){
int array[]={10,20,30,40};
printf("%d",-2[array]);
}

(A) -60
(B) -30
(C) 60
(D) Garbage value
(E) Compiler error


Explanation:
In c,
array[2]=*(array+2)=*(2+array)=2[array]=30

18.

What will be output if you will compile and execute the following c code?

void main(){
int i=10;
static int x=i;
if(x==i)
printf("Equal");
else if(x>i)
printf("Greater than");
else
printf("Less than");
}

(A) Equal
(B) Greater than
(C) Less than
(D) Compiler error
(E) None of above


Explanation:
static variables are load time entity while auto variables are run time entity. We can not initialize any load time variable by the run time variable.
In this example i is run time variable while x is load time variable.
What is storage class?

19.

What will be output if you will compile and execute the following c code?

#define max 5;
void main(){
int i=0;
i=max++;
printf("%d",i++);
}

(A) 5
(B) 6
(C) 7
(D) 0
(E) Compiler error


Explanation:
#define is token pasting preprocessor. If you will see intermediate file: test.i
test.c 1:
test.c 2: void main(){
test.c 3: int i=0;
test.c 4: i=5++;
test.c 5: printf("%d",i++);
test.c 6: }
test.c 7:
It is clear macro constant max has replaced by 5. It is illegal to increment the constant number. Hence compiler will show Lvalue required.
Preprocessor questions and answer.

20.

What will be output if you will compile and execute the following c code?

void main(){
double far* p,q;
printf("%d",sizeof(p)+sizeof q);
}

(A) 12
(B) 8
(C) 4
(D) 1
(E) Compiler error


Explanation:

It is clear p is far pointer and size of far pointer is 4 byte while q is double variable and size of double variable is 8 byte.



(1) What will be output if you will compile and execute the following c code?

struct marks{
  int p:3;
  int c:3;
  int m:2;
};
void main(){
  struct marks s={2,-6,5};
  printf("%d %d %d",s.p,s.c,s.m);
}

(a) 2 -6 5
(b) 2 -6 1
(c) 2 2 1
(d) Compiler error
(e) None of these

Answer: (c)
Explanation:
Binary value of 2: 00000010 (Select three two bit)
Binary value of 6: 00000110
Binary value of -6: 11111001+1=11111010
(Select last three bit)
Binary value of 5: 00000101 (Select last two bit)

Complete memory representation:


(2) What will be output if you will compile and execute the following c code?

void main(){
   int huge*p=(int huge*)0XC0563331;
   int huge*q=(int huge*)0xC2551341;
   *p=200;
   printf("%d",*q);
}

(a)0
(b)Garbage value
(c)null
(d) 200
(e)Compiler error

Answer: (d)
Explanation:
Physical address of huge pointer p
Huge address: 0XC0563331
Offset address: 0x3331
Segment address: 0XC056
Physical address= Segment address * 0X10 + Offset address
=0XC056 * 0X10 +0X3331
=0XC0560 + 0X3331
=0XC3891
Physical address of huge pointer q
Huge address: 0XC2551341
Offset address: 0x1341
Segment address: 0XC255
Physical address= Segment address * 0X10 + Offset address
=0XC255 * 0X10 +0X1341
=0XC2550 + 0X1341
=0XC3891
Since both huge pointers p and q are pointing same physical address so content of q will also same as content of q.

(3) Write c program which display mouse pointer and position of pointer.(In x coordinate, y coordinate)?

Answer:
#include”dos.h”
#include”stdio.h”
void main()
{
union REGS i,o;
int x,y,k;
//show mouse pointer
i.x.ax=1;
int86(0x33,&i,&o);
while(!kbhit()) //its value will false when we hit key in the key board
{
i.x.ax=3; //get mouse position
x=o.x.cx;
y=o.x.dx;
clrscr();
printf("(%d , %d)",x,y);
delay(250);
int86(0x33,&i,&o);
}
getch();
}

(4) Write a c program to create dos command: dir.

Answer:

Step 1: Write following code.

#include “stdio.h”
#include “dos.h”
void main(int count,char *argv[]){
   struct find_t q ;
   int a;
   if(count==1)
      argv[1]="*.*";
      a = _dos_findfirst(argv[1],1,&q);
      if(a==0){
         while (!a){
            printf(" %s\n", q.name);
            a = _dos_findnext(&q);
         }
      }
      else{
         printf("File not found");
      }
}

Step 2: Save the as list.c (You can give any name)
Step 3: Compile and execute the file.
Step 4: Write click on My computer of Window XP operating system and select properties.
Step 5: Select Advanced -> Environment Variables
Step 6: You will find following window:
Click on new button (Button inside the red box)




Step 7: Write following:
Variable name: path
Variable value: c:\tc\bin\list.c (Path where you have saved)


Step 8: Open command prompt and write list and press enter.
Command line argument tutorial.

(6) What will be output if you will compile and execute the following c code?

void main(){
   int i=10;
   static int x=i;
   if(x==i)
      printf("Equal");
   else if(x>i)
      printf("Greater than");
   else
      printf("Less than");
}

(a) Equal
(b) Greater than
(c) Less than
(d) Compiler error
(e) None of above

Answer: (d)
Explanation:
static variables are load time entity while auto variables are run time entity. We can not initialize any load time variable by the run time variable.
In this example i is run time variable while x is load time variable.


(7) What will be output if you will compile and execute the following c code?

void main(){
   int i;
   float a=5.2;
   char *ptr;
   ptr=(char *)&a;
   for(i=0;i<=3;i++)
      printf("%d ",*ptr++);
}

(a)0 0 0 0
(b)Garbage Garbage Garbage Garbage
(c)102 56 -80 32
(d)102 102 -90 64
(e)Compiler error

Answer: (d)
Explanation:
In c float data type is four byte data type while char pointer ptr can point one byte of memory at a time.
Memory representation of float a=5.2




ptr pointer will point first fourth byte then third byte then second byte then first byte.
Content of fourth byte:
Binary value=01100110
Decimal value= 64+32+4+2=102
Content of third byte:
Binary value=01100110
Decimal value=64+32+4+2=102
Content of second byte:
Binary value=10100110
Decimal value=-128+32+4+2=-90
Content of first byte:
Binary value=01000000
Decimal value=64
Note: Character pointer treats MSB bit of each byte i.e. left most bit of above figure as sign bit.

(8) What will be output if you will compile and execute the following c code?

void main(){
   int i;
   double a=5.2;
   char *ptr;
   ptr=(char *)&a;
   for(i=0;i<=7;i++)
      printf("%d ",*ptr++);
}

(a) -51 -52 -52 -52 -52 -52 20 64
(b) 51 52 52 52 52 52 20 64
(c) Eight garbage values.
(d) Compiler error
(e) None of these

Answer: (a)
Explanation:
In c double data type is eight byte data type while char pointer ptr can point one byte of memory at a time.
Memory representation of double a=5.2




ptr pointer will point first eighth byte then seventh byte then sixth byte then fifth byte then fourth byte then third byte then second byte then first byte as shown in above figure.
Content of eighth byte:
Binary value=11001101
Decimal value= -128+64+8+4+1=-51
Content of seventh byte:
Binary value=11001100
Decimal value= -128+64+8+4=-52
Content of sixth byte:
Binary value=11001100
Decimal value= -128+64+8+4=-52
Content of fifth byte:
Binary value=11001100
Decimal value= -128+64+8+4=-52
Content of fourth byte:
Binary value=11001100
Decimal value= -128+64+8+4=-52
Content of third byte:
Binary value=11001100
Decimal value= -128+64+8+4=-52
Content of second byte:
Binary value=000010100
Decimal value=16+4=20
Content of first byte:
Binary value=01000000
Decimal value=64
Note: Character pointer treats MSB bit of each byte i.e. left most bit of above figure as sign bit.

(9) What will be output if you will compile and execute the following c code?

void main(){
   printf("%s","c" "question" "bank");
}

(a) c question bank
(b) c
(c) bank
(d) cquestionbank
(e) Compiler error

Answer: (d)
Explanation:
In c string constant “xy” is same as “x” “y”


(10) What will be output if you will compile and execute the following c code?

void main(){
   printf("%s",__DATE__);
}

(a) Current system date
(b) Current system date with time
(c) null
(d) Compiler error
(e) None of these

Answer: (a)
Explanation:
__DATE__ is global identifier which returns current system date.

(11) What will be output if you will compile and execute the following c code?

void main(){
   char *str="c-pointer";
   printf("%*.*s",10,7,str);
}

(a) c-pointer
(b) c-pointer
(c) c-point
(d) cpointer null null
(e) c-point

Answer: (e)
Explanation:
Meaning of %*.*s in the printf function:
First * indicates the width i.e. how many spaces will take to print the string and second * indicates how many characters will print of any string.
Following figure illustrates output of above code:




(12) What will be output if you will compile and execute the following c code?

void start();
void end();
#pragma startup start
#pragma exit end
int static i;
void main(){
   printf("\nmain function: %d",++i);
}
void start(){
   clrscr();
   printf("\nstart function: %d",++i);
}
void end(){
   printf("\nend function: %d",++i);
   getch();
}

(a)
main function: 2
start function: 1
end function:3
(b)
start function: 1
main function: 2
end function:3
(c)
main function: 2
end function:3
start function: 1
(d) Compiler error
(e) None of these

Answer: (b)
Explanation:
Every c program start with main function and terminate with null statement. But #pragma startup can call function just before main function and #pragma exit
(13) What will be output if you will compile and execute the following c code?

void main(){
   int a=-12;
   a=a>>3;
   printf("%d",a);
}

(a) -4
(b) -3
(c) -2
(d) -96
(e) Compiler error

Answer :( c)
Explanation:
Binary value of 12 is: 00000000 00001100
Binary value of -12 wills 2’s complement of 12 i.e.




So binary value of -12 is: 11111111 11110100




Right shifting rule:
Rule 1: If number is positive the fill vacant spaces in the left side by 0.
Rule 2: If number is negative the fill vacant spaces in the left side by 1.
In this case number is negative. So right shift all the binary digits by three space and fill vacant space by 1 as shown following figure:




Since it is negative number so output will also a negative number but its 2’s complement.




Hence final out put will be:




And its decimal value is: 2
Hence output will be:-2

(14) What will be output if you will compile and execute the following c code?

#include "string.h"
void main(){
   clrscr();
 printf("%d%d",sizeof("string"),strlen("string"));
getch();
}
(a) 6 6
(b) 7 7
(c) 6 7
(d) 7 6
(e) None of these

Answer: (d)
Explanation:
Sizeof operator returns the size of string including null character while strlen function returns length of a string excluding null character.

(15) What will be output if you will compile and execute the following c code?

void main(){
   static main;
   int x;
   x=call(main);
   clrscr();
   printf("%d ",x);
   getch();
}
int call(int address){
   address++;
   return address;
}
(a) 0
(b) 1
(c) Garbage value
(d) Compiler error
(e) None of these

Answer: (b)
Explanation:
As we know main is not keyword of c but is special type of function. Word main can be name variable in the main and other functions.
(16) What will be output if you will compile and execute the following c code?

void main(){
   int a,b;
   a=1,3,15;
   b=(2,4,6);
   clrscr();
   printf("%d ",a+b);
   getch();
}
(a) 3
(b) 21
(c) 17
(d) 7
(e) Compiler error

Answer: (d)
Explanation:
In c comma behaves as separator as well as operator.
a=1, 3, 15;
b= (2, 4, 6);
In the above two statements comma is working as operator. Comma enjoys least precedence and associative is left to right.
Assigning the priority of each operator in the first statement:




Hence 1 will assign to a.
Assigning the priority of each operator in the second statement:



(17) What will be output if you will compile and execute the following c code?

int extern x;
void main()
   printf("%d",x);
   x=2;
   getch();
}
int x=23;

(a) 0
(b) 2
(c) 23
(d) Compiler error
(e) None of these

Answer: (c)
Explanation:
extern variables can search the declaration of variable any where in the program.

(18) What will be output if you will compile and execute the following c code?

void main(){
   int i=0;
   if(i==0){
      i=((5,(i=3)),i=1);
      printf("%d",i);
   }
   else
      printf("equal");
}

(a) 5
(b) 3
(c) 1
(d) equal
(e) None of above

Answer: (c)
Explanation:

(19) What will be output if you will compile and execute the following c code?

void main(){
   int a=25;
   clrscr();
   printf("%o %x",a,a);
   getch();
}

(a) 25 25
(b) 025 0x25
(c) 12 42
(d) 31 19
(e) None of these

Answer: (d)
Explanation:
%o is used to print the number in octal number format.
%x is used to print the number in hexadecimal number format.
Note: In c octal number starts with 0 and hexadecimal number starts with 0x.

(20) What will be output if you will compile and execute the following c code?

#define message "union is\
power of c"
void main(){
   clrscr();
   printf("%s",message);
   getch();
}

(a) union is power of c
(b) union ispower of c
(c) union is
Power of c
(d) Compiler error
(e) None of these

Answer: (b)
Explanation:
If you want to write macro constant in new line the end with the character \.

(21) What will be output if you will compile and execute the following c code?

#define call(x) #x
void main(){
   printf("%s",call(c/c++));
}

(a)c
(b)c++
(c)#c/c++
(d)c/c++
(e)Compiler error

Answer: (d)
Explanation:
# is string operator. It converts the macro function call argument in the string. First see the intermediate file:
test.c 1:
test.c 2: void main(){
test.c 3: printf("%s","c/c++");
test.c 4: }
test.c 5:
It is clear macro call is replaced by its argument in the string format.

(22) What will be output if you will compile and execute the following c code?

void main(){
   if(printf("cquestionbank"))
      printf("I know c");
   else
      printf("I know c++");
}

(a) I know c
(b) I know c++
(c) cquestionbankI know c
(d) cquestionbankI know c++
(e) Compiler error

Answer: (c)
Explanation:
Return type of printf function is integer which returns number of character it prints including blank spaces. So printf function inside if condition will return 13. In if condition any non- zero number means true so else part will not execute.


No comments:

Post a Comment