1. What is the difference between Swapping and Paging?
Swapping:
Whole process is moved from the swap device to the main memory for
execution. Process size must be less than or equal to the available main memory. It is easier to implementation and overhead to the system. Swapping systems does not handle the memory more flexibly as compared to the paging systems.
Paging:
Only the required memory pages are moved to main memory from the swap device for execution. Process size does not matter. Gives the concept of the virtual memory.
2. Is the Process before and after the swap are the same? Give reason.
Process before swapping is residing in the primary memory in its original form. The regions (text, data and stack) may not be occupied fully by the process, there may be few empty slots in any of the regions and while swapping Kernel do not bother about the empty slots while swapping the process out.
After swapping the process resides in the swap (secondary memory) device. The regions swapped out will be present but only the occupied region slots but not the empty slots that were present before assigning.
While swapping the process once again into the main memory, the Kernel referring to the Process Memory Map, it assigns the main memory accordingly taking care of the empty slots in the regions.
3. What is a Map?
A Map is an Array, which contains the addresses of the free space in the swap device that are allocatable resources, and the number of the resource units available there.
This allows First-Fit allocation of contiguous blocks of a resource. Initially the Map contains one entry – address (block offset from the starting of the swap area) and the total number of resources.
Kernel treats each unit of Map as a group of disk blocks. On the allocation and freeing of the resources Kernel updates the Map for accurate information.
4. What are the requirements for a swapper to work?
The swapper works on the highest scheduling priority. Firstly it will look for any sleeping process, if not found then it will look for the ready-to-run process for swapping. But the major requirement for the swapper to work the ready-to-run process must be core-resident for at least 2 seconds before swapping out. And for swapping in the process must have been resided in the swap device for at least 2 seconds. If the requirement is not satisfied then the swapper will go into the wait state on that event and it is awaken once in a second by the Kernel.
5. What are the criteria for choosing a process for swapping into memory from the swap device?
The resident time of the processes in the swap device, the priority of the processes and the amount of time the processes had been swapped out.
6. What are the criteria for choosing a process for swapping out of the memory to the swap device?
>The process’s memory resident time,
>Priority of the process and
>The nice value.
7. What do you mean by nice value?
Nice value is the value that controls {increments or decrements} the priority of the process. This value that is returned by the nice () system call. The equation for using nice value is:
Priority = (“recent CPU usage”/constant) + (base- priority) + (nice value) 130
8. What is Page-Stealer process?
This is the Kernel process that makes rooms for the incoming pages, by swapping the memory pages that are not the part of the working set of a process. Page-Stealer is created by the Kernel at the system initialization and invokes it throughout the lifetime of the system. Kernel locks a region when a process faults on a page in the region, so that page stealer cannot steal the page, which is being faulted in.
9. What is validity fault?
If a process referring a page in the main memory whose valid bit is not set, it results in validity fault.
The valid bit is not set for those pages:
>that are outside the virtual address space of a process,
>that are the part of the virtual address space of the process but no physical address is assigned to it.
10. What do you mean by the protection fault?
Protection fault refers to the process accessing the pages, which do not have the access permission. A process also incur the protection fault when it attempts to write a page whose copy on write bit was set during the fork() system call.
11. What are links and symbolic links in UNIX file system?
A link is a second name (not a file) for a file. Links can be used to assign more than one name to a file, but cannot be used to assign a directory more than one name or link filenames on different computers.
Symbolic link 'is' a file that only contains the name of another file. Operation on the symbolic link is directed to the file pointed by it. Both the limitations of links are eliminated in symbolic links.
Commands for linking files are:
Link ln filename1 filename2
Symbolic link ln -s filename1 filename2
12. What is meant by arm-stickiness?
If one or a few processes have a high access rate to data on one track of a storage disk, then they may monopolize the device by repeated requests to that track. This generally happens with most common device scheduling algorithms (LIFO, SSTF, C-SCAN, etc). High-density multisurface disks are more likely to be affected by this than low density ones
Sunday, March 21, 2010
Help for C
1. what is the output of the program?
#include
main()
{
struct s1 {int i; };
struct s2 {int i; };
struct s1 st1;
struct s2 st2;
st1.i =5;
st2 = st1;
printf(" %d " , st2.i);
}
ans: nothing (error)
expl: diff struct variables should not assigned using "=" operator.
2. what is the output of the program?
main()
{
struct emp{
char emp[];
int empno;
float sal;
};
struct emp member = { "TIGER"};
printf(" %d %f", member.empno,member.sal);
ans: error. In struct variable emp[], we have to give array size.
If array size given
ans is 0, 0.00
3. if ptr is defined as
int *ptr[][100];
which of the following correctly allocates memory for ptr?
ans: ptr = (int *)(malloc(100* sizeof(int));
4. What is the output for the program given below
typedef enum grade{GOOD,BAD,WORST,}BAD;
main()
{
BAD g1;
g1=1;
printf("%d",g1);
}
4. Give the output for the following program.
#define STYLE1 char
main()
{
typedef char STYLE2;
STYLE1 x;
STYLE2 y;
clrscr();
x=255;
y=255;
printf("%d %d\n",x,y);
}
5. main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.
6. main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.
7. {
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.
8. {
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer: Ok here
Explanation:
Printf will return how many characters does it print. Hence printing
a new-line character returns 1 which makes the if statement true, thus "Ok here" is printed.
9. void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.
10. int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.
11. main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
12. #include
main()
{
char * str = "hello";
char * ptr = str;
char least = 127; 65
while(*ptr++)
least = (*ptr
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.
13. Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.
14. main
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
15. main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string “HELL” and doesn’t have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.
16. main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
}
int i;
17. char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.
18. char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.
19. main()
{
char c=-64;
int i=-32;
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i
printf("pass2");
else
printf("Fail2")
}
20. int i,j,k;
i=2;
j=4;
k=i++>j&&2; /* (i++>j)&2 */
printf("%d\n",k);
if(++k && ++i<--j|| i++)
j=++k;
printf(" %d %d %d",i,-j--,k);
21. void main()
{
char *st1[3]= {"Hello","World","Oracle"};
*st1=st1[2];
st1[1]=*st1;
free(st1[0]);
free(st1[1]);
clrscr();
printf("%s %s %s",st1,st1[1],st1[2]); //Garbeg Oracle Oracle
}
22. main()
{
char c = 255;printf("%d",c);
}
OUTPUT:-1
23. main()
{
int i,count=0;
char *p1="abcdefghij";
char *p2="alcmenfoip";
for(i=0;i<=strlen(p1);i++) // strlen gets decremented with p++
{
if(*p1++ == *p2++) // loop runs till i=5
count+=5;
else
count-=3;
}
printf("count=%d\n",count);
}
24. #define mysizeof(a) (&a+1) - &a
main()
{
float d;
printf("%d\n", mysizeof(d) );
}
note: assume sizeof float is 8 bytes
a) 8
b) 4
c) 1
d) compiler error
25. int A=1,B=2;
if(A==B < printf("Hello "))
printf("world\n");
else
printf("Bangalore\n");
What is the o/p?
a> world
b> Hello bangalore
c> bangalore
d> Hello world.
26. register int a,b; //ERROR
main()
{
for(a=0 ; a<5 ; a++)
b++;
}
27. main()
{
for( printf("a") ; printf("b") ; printf("c") ) ;
}
a) abc
b) abc abc abc .....(infinite times)
c) a bc bc bc ....(infinite times)
d) Error
28. main()
{
int i = 100 ;
printf("%d ", sizeof(i++));
printf("%d ",i) ;
}
Ans.4,100
29. union tag
{
int a;
char x;
char y;
}name;
int main()
{
name.a=5;
printf("\n x = %d y = %d ",name.x,name.y); // OUTPUT : 5 5
}
30. main()
{
#define x 5 //NO ERROR
int b;
b = x;
printf("%d",b);
}
31. main()
{
int a; #define y 10 //ERROR
a=y;
printf("%d",a);
}
32. struct
{
int i;
}node ;
main()
{
printf("%d",node.i);
}
(a). 0
(b). Garbage value
(c). error.
(d). warning
33. i=5;
i=i++ * i++;
printf("%d",i);
a)30
b)49
c)25
d)27
34. i=5;
printf("%d",i++ * i++);
a)30
b)49
c)25
d)37
35. main()
{
int *p,*q,i;
p=(int *)100;
q=(int *)200;
i=q-p;
printf("%d",i);
}
a)100 b)25 c)0 d)compile error
36. main()
{
int a=10,b=5;
if(a=a&b)
b=a^b;
printf("a=%d,b=%d",a,b);
}
a)a=0,b=5
b)a=10 b=5
37. int main()
{
int i=89;
int *p; char *t;
p=&i;
t=(char *)&i;
if( *p==*t)
printf("yes"); //Ans
else
printf("no");
}
#include
main()
{
struct s1 {int i; };
struct s2 {int i; };
struct s1 st1;
struct s2 st2;
st1.i =5;
st2 = st1;
printf(" %d " , st2.i);
}
ans: nothing (error)
expl: diff struct variables should not assigned using "=" operator.
2. what is the output of the program?
main()
{
struct emp{
char emp[];
int empno;
float sal;
};
struct emp member = { "TIGER"};
printf(" %d %f", member.empno,member.sal);
ans: error. In struct variable emp[], we have to give array size.
If array size given
ans is 0, 0.00
3. if ptr is defined as
int *ptr[][100];
which of the following correctly allocates memory for ptr?
ans: ptr = (int *)(malloc(100* sizeof(int));
4. What is the output for the program given below
typedef enum grade{GOOD,BAD,WORST,}BAD;
main()
{
BAD g1;
g1=1;
printf("%d",g1);
}
4. Give the output for the following program.
#define STYLE1 char
main()
{
typedef char STYLE2;
STYLE1 x;
STYLE2 y;
clrscr();
x=255;
y=255;
printf("%d %d\n",x,y);
}
5. main()
{
while (strcmp(“some”,”some\0”))
printf(“Strings are not equal\n”);
}
Answer:
No output
Explanation:
Ending the string constant with \0 explicitly makes no difference. So “some” and “some\0” are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.
6. main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’\0’};
while (strcmp(str1,str2))
printf(“Strings are not equal\n”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.
7. {
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.
8. {
char a[]="\0";
if(printf("%s\n",a))
printf("Ok here \n");
else
printf("Forget it\n");
}
Answer: Ok here
Explanation:
Printf will return how many characters does it print. Hence printing
a new-line character returns 1 which makes the if statement true, thus "Ok here" is printed.
9. void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.
10. int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.
11. main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.
12. #include
main()
{
char * str = "hello";
char * ptr = str;
char least = 127; 65
while(*ptr++)
least = (*ptr
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.
13. Is there any difference between the two declarations,
1. int foo(int *arr[]) and
2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are allowed inside the [] is just for more readability. So there is no difference between the two declarations.
14. main
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
15. main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string “HELL” and doesn’t have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.
16. main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
}
int i;
17. char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.
18. char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.
19. main()
{
char c=-64;
int i=-32;
unsigned int u =-16;
if(c>i)
{printf("pass1,");
if(c
printf("pass2");
else
printf("Fail2");
}
else
printf("Fail1);
if(i
printf("pass2");
else
printf("Fail2")
}
20. int i,j,k;
i=2;
j=4;
k=i++>j&&2; /* (i++>j)&2 */
printf("%d\n",k);
if(++k && ++i<--j|| i++)
j=++k;
printf(" %d %d %d",i,-j--,k);
21. void main()
{
char *st1[3]= {"Hello","World","Oracle"};
*st1=st1[2];
st1[1]=*st1;
free(st1[0]);
free(st1[1]);
clrscr();
printf("%s %s %s",st1,st1[1],st1[2]); //Garbeg Oracle Oracle
}
22. main()
{
char c = 255;printf("%d",c);
}
OUTPUT:-1
23. main()
{
int i,count=0;
char *p1="abcdefghij";
char *p2="alcmenfoip";
for(i=0;i<=strlen(p1);i++) // strlen gets decremented with p++
{
if(*p1++ == *p2++) // loop runs till i=5
count+=5;
else
count-=3;
}
printf("count=%d\n",count);
}
24. #define mysizeof(a) (&a+1) - &a
main()
{
float d;
printf("%d\n", mysizeof(d) );
}
note: assume sizeof float is 8 bytes
a) 8
b) 4
c) 1
d) compiler error
25. int A=1,B=2;
if(A==B < printf("Hello "))
printf("world\n");
else
printf("Bangalore\n");
What is the o/p?
a> world
b> Hello bangalore
c> bangalore
d> Hello world.
26. register int a,b; //ERROR
main()
{
for(a=0 ; a<5 ; a++)
b++;
}
27. main()
{
for( printf("a") ; printf("b") ; printf("c") ) ;
}
a) abc
b) abc abc abc .....(infinite times)
c) a bc bc bc ....(infinite times)
d) Error
28. main()
{
int i = 100 ;
printf("%d ", sizeof(i++));
printf("%d ",i) ;
}
Ans.4,100
29. union tag
{
int a;
char x;
char y;
}name;
int main()
{
name.a=5;
printf("\n x = %d y = %d ",name.x,name.y); // OUTPUT : 5 5
}
30. main()
{
#define x 5 //NO ERROR
int b;
b = x;
printf("%d",b);
}
31. main()
{
int a; #define y 10 //ERROR
a=y;
printf("%d",a);
}
32. struct
{
int i;
}node ;
main()
{
printf("%d",node.i);
}
(a). 0
(b). Garbage value
(c). error.
(d). warning
33. i=5;
i=i++ * i++;
printf("%d",i);
a)30
b)49
c)25
d)27
34. i=5;
printf("%d",i++ * i++);
a)30
b)49
c)25
d)37
35. main()
{
int *p,*q,i;
p=(int *)100;
q=(int *)200;
i=q-p;
printf("%d",i);
}
a)100 b)25 c)0 d)compile error
36. main()
{
int a=10,b=5;
if(a=a&b)
b=a^b;
printf("a=%d,b=%d",a,b);
}
a)a=0,b=5
b)a=10 b=5
37. int main()
{
int i=89;
int *p; char *t;
p=&i;
t=(char *)&i;
if( *p==*t)
printf("yes"); //Ans
else
printf("no");
}
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