(1) What will be output of following program?void main()
{
int a=320;
char *ptr;
ptr=(char *)&a;
printf("%d ",*ptr);
getch();
}
(a) 2
(b) 320
(c) 64
(d)Compiler error
(e)None of above
(2) What will be output of following program?#include"stdio.h"
#include"conio.h"
void main()
{
void (*p)();
int (*q)();
int (*r)();
p=clrscr;
q=getch;
r=puts;
(*p)();
(*r)("cquestionbank.blogspot.com");
(*q)();
}
(a) NULL
(b) cquestionbank.blogspot.com
(c) c
(d) Compiler error
(e) None of above
(3) What will be output of following program?
void main()
{
int i=3;
int *j;
int **k;
j=&i;
k=&j;
printf(“%u %u %d ”,k,*k,**k);
}
(a) Address, Address, 3
(b)Address, 3, 3
(c) 3, 3, 3
(d) Compiler error
(e) None of above
(4) What will be output of following program?void main()
{
char far *p=(char far *)0x55550005;
char far *q=(char far *)0x53332225;
*p=80;
(*p)++;
printf("%d",*q);
getch();
}
(a) 80
(b) 81
(c) 82
(d) Compiler error
(e) None of above
(5) What will be output of following program?
#include"stdio.h"
#include"string.h"
void main()
{
char *ptr1=NULL;
char *ptr2=0;
strcpy(ptr1," c");
strcpy(ptr2,"questions");
printf("\n%s %s",ptr1,ptr2);
getch();
}
(a) c questions
(b) c (null)
(c) (null) (null)
(d) Compiler error
(e) None of above
(6) What will be output of following program?void main()
{
int huge *a=(int huge *)0x59990005;
int huge *b=(int huge *)0x59980015;
if(a==b)
printf("power of pointer");
else
printf("power of c");
getch();
}
(a) power of pointer
(b) power of c
(c) power of cpower of c
(d) Compiler error
(e) None of above
(7) What will be output of following program?
#include"stdio.h"
#include"string.h"
void main()
{
register a=25;
int far *p;
p=&a;
printf("%d ",*p);
getch();
}
(a) 25
(b) 4
(c) Address
(d) Compiler error
(e) None of above
(8) What will be output of following program?#include"stdio.h"
#include"string.h"
void main()
{
char far *p,*q;
printf("%d %d",sizeof(p),sizeof(q));
getch();
}
(a) 2 2
(b) 4 4
(c) 4 2
(d) 2 4
(e) None of above
(9) What will be output of following program?void main()
{
int a=10;
void *p=&a;
int *ptr=p;
printf("%u",*ptr);
getch();
}
(a) 10
(b) Address
(c) 2
(d) Compiler error
(e) None of above
(10) What will be output of following program?#include"stdio.h"
#include"string.h"
void main()
{
int register a;
scanf("%d",&a);
printf("%d",a);
getch();
}
//if a=25
(a) 25
(b) Address
(c) 0
(d) Compiler error
(e) None of above
(11) What will be output of following program?void main()
{
char arr[10];
arr="world";
printf("%s",arr);
getch();
}
(a) world
(b) w
(c) Null
(d) Compiler error
(e) None of above
(12) What will be output of following program?
#include"stdio.h"
#include"string.h"
void main()
{
int a,b,c,d;
char *p=0;
int *q=0;
float *r=0;
double *s=0;
a=(int)(p+1);
b=(int)(q+1);
c=(int)(r+1);
d=(int)(s+1);
printf("%d %d %d %d",a,b,c,d);
}
(a)2 2 2 2
(b)1 2 4 8
(c)1 2 2 4
(d) Compiler error
(e) None of above
(13) What will be output of following program?
#include"stdio.h"
#include"string.h"
void main()
{
int a=5,b=10,c;
int *p=&a,*q=&b;
c=p-q;
printf("%d",c);
getch();
}
(a) 1
(b) 5
(c) -5
(d) Compiler error
(e) None of above
(14) What will be output of following program?
unsigned long int (* avg())[3]
{
static unsigned long int arr[3]={1,2,3};
return &arr;
}
void main()
{
unsigned long int (*ptr)[3];
ptr=avg();
printf("%d",*(*ptr+2));
getch();
}
(a) 1
(b) 2
(c) 3
(d) Compiler error
(e) None of above
(15) What will be output of following program?
void main()
{
int * p,b;
b=sizeof(p);
printf(“%d”,b);
}
(a) 2
(b) 4
(c) 8
(d) Compiler error
(e) None of above
(16) What will be output of following program?
void main()
{
int i=5,j;
int *p,*q;
p=&i;
q=&j;
j=5;
printf("value of i : %d value of j : %d",*p,*q);
getch();
}
(a) 5 5
(b) Address Address
(c) 5 Address
(d) Compiler error
(e) None of above
(17) What will be output of following program?
{
int i=5;
int *p;
p=&i;
printf(" %u %u",*&p,&*p);
getch();
}
(a) 5 Address
(b) Address Address
(c) Address 5
(d) Compiler error
(e) None of above
(18) What will be output of following program? void main()
{
int i=100;
printf("value of i : %d addresss of i : %u",i,&i);
i++;
printf("\nvalue of i : %d addresss of i : %u",i,&i);
getch();
}
(a) value of i : 100 addresss of i : Address
value of i : 101 addresss of i : Address
(b) value of i : 100 addresss of i : Address
value of i : 100 addresss of i : Address
(c) value of i : 101 addresss of i : Address
value of i : 101 addresss of i : Address
(d) Compiler error
(e) None of above
(19) What will be output of following program? void main()
{
char far *p=(char far *)0x55550005;
char far *q=(char far *)0x53332225;
*p=25;
(*p)++;
printf("%d",*q);
getch();
}
(a) 25
(b) Address
(c) Garbage
(d) Compiler error
(e) None of above
(20) What will be output of following program? void main()
{
int i=3;
int *j;
int **k;
j=&i;
k=&j;
printf(“%u %u %u”,i,j,k);
}
(a) 3 Address 3
(b) 3 Address Address
(c) 3 3 3
(d) Compiler error
(e) None of above
Answer:
(1) c
(2) b
(3) a
(4) b
(5) c
(6) a
(7) d
(8) c
(9) a
(10) d
(11) d
(12) b
(13) a
(14) c
(15) e
(16) a
(17) b
(18) a
(19) e
(20) b
Explanation
(1)
As we know int is two byte data byte while char is one byte data byte. char pointer can keep the address one byte at time.
Binary value of 320 is 00000001 01000000 (In 16 bit)
Memory representation of int a=320 is:
So ptr is pointing only first 8 bit which color is green and
Decimal value is 64.
(2)
p is pointer to function whose parameter is void and return type
is also void. r and q is pointer to function whose parameter is
void and return type is int . So they can hold the address of
such function.
(3)
Memory representation
Here 6024, 8085, 9091 is any arbitrary address, it may be different.
Value of k is content of k in memory which is 8085
Value of *k means content of memory location which address k keeps.
k keeps address 8085 .
Content of at memory location 8085 is 6024
In the same way **k will equal to 3.
Short cut way to calculate:
Rule: * and & always cancel to each other
i.e. *&a=a
So *k=*(&j) since k=&j
*&j=j =6024
And
**k=**(&j)=*(*&j)=*j=*(&i)=*&i=i=3
(4)
Far address of p and q are representing same physical address .
Physical address of 0x55550005= (0x5555)*(0x10) + (0x0005) = 0x55555
Physical address of 0x53332225= (0x5333*0x10) + (0x2225) =0x55555
*p =80, means content at memory location 0x55555 is assigning value 25
(*p)++ means increase the content by one at memory location 0x5555 so now
content at memory location 0x55555 is 81
*q also means content at memory location 0x55555 which is 26
(5)
We cannot assign any string constant in null pointer by strcpy function.
(6)
Here we are performing relational operation between two huge addresses. So
first both a and b will normalize.
a= (0x5999)* (0x10) + (0x0005) =0x9990+0x0005=0x9995
b= (0x5998)* (0x10) + (0x0015) =0x9980+0x0015=0x9995
Here both huge addresses are representing same physical address. So a==b
is true.
(7)
Register data type stores in CPU. So it has not any memory
address. Hence we cannot write &a.
(8)
p is far pointer which size is 4 byte.
By default q is near pointer which size is 2 byte.
(9)
void pointer can hold address of any data type without type casting. Any
pointer can hold void pointer without type casting.
(10)
Register data type stores in CPU. So it has not any memory address. Hence
we cannot write &a.
(11) Compiler error Lvalue required
Array name is constant pointer and we cannot assign any
value in constant data type after declaration.
(12)
Address=next address
Since initial address of all data type is zero. So its
next address will be size of data type.
(13)
Difference of two same type of pointer is always one.
(15)
Output: 2 or 4
since in this question it has not written p is which type
of pointer. So its output will depend upon which memory
model has selected. Default memory model is small.
(17)
Since * and & always cancel to each other.
i.e. *&a=a
so *&p=p which store address of integer i
&*p=&*(&i) //since p=&i
=&(*&i)
=&i
So second output is also address of i
(18)
Within the scope of any variable, value of variable
may change but its address will never change in any
modification of variable.
(19)
Far address of p and q are representing same physical
address. Physical address of
0x55550005= 0x5555*ox10+ox0005= 0x55555
Physical address of
0x53332225=0x5333*0x10+ox2225=0x55555
*p =25, means content at memory location 0x55555 is
assigning value 25
(*p)++ means increase the content by one at memory
location 0x5555 so now content at memory location 0x55555
is 26
*q also means content at memory location 0x55555 which is
26
(20)
Here 6024, 8085, 9091 is any arbitrary address, it may be different.
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