Poornam Info Vision C questions with answers

Poornam Info Vision C questions with answers

 

C Questions

 

1Predict the output or error(s) for the following:

void main()

{

int const * p=5;

printf(“%d”,++(*p));

}

Answer:

Compiler error: Cannot modify a constant value.

Explanation: p is a pointer to a “constant integer”. But we tried to change the value of the “constant integer”.

 

  1. main()

{

char s[ ]=”man”;

int i;

for(i=0;s[ i ];i++)

printf(“\n%c%c%c%c”,s[ i ],*(s+i),*(i+s),i[s]);

}

Answer:

mmmm

aaaa

nnnn

Explanation: s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index  number/ displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

 

  1. main()

{

float me = 1.1;

double you = 1.1;

if(me==you)

printf(“I love U”);

else

printf(“I hate U”);

}

Answer: I hate U

 

Explanation:  For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Rule of Thumb:

Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

 

  1. main()

{

static int var = 5;

printf(“%d “,var–);

if(var)

main();

}

Answer: 5 4 3 2 1

Explanation: When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

 

  1. main()

{

int c[ ]={2.8,3.4,4,6.7,5};

int j,*p=c,*q=c;

for(j=0;j<5;j++) {

printf(” %d “,*c);

++q; }

for(j=0;j<5;j++){

printf(” %d “,*p);

++p; }

}

Answer:

2 2 2 2 2 2 3 4 6 5

Explanation: Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

 

  1.  main()

{

extern int i;

i=20;

printf(“%d”,i);

}

Answer:

Linker Error : Undefined symbol ‘_i’

Explanation: extern storage class in the following declaration,

extern int i;

specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

 

  1.  main()

{

int i=-1,j=-1,k=0,l=2,m;

m=i++&&j++&&k++||l++;

printf(“%d %d %d %d %d”,i,j,k,l,m);

}

Answer: 0 0 1 3 1

Explanation: Logical operations always give a result of 1 or 0. And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

 

  1.  main()

{

char *p;

printf(“%d %d “,sizeof(*p),sizeof(p));

}

Answer: 1 2

Explanation: The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

 

  1.  main()

{

int i=3;

switch(i)

{

default:printf(“zero”);

case 1: printf(“one”);

break;

case 2:printf(“two”);

break;

case 3: printf(“three”);

break;

}

}

Answer : Three

Explanation: The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn’t match.

 

  1.  main()

{

printf(“%x”,-1<<4);

}

Answer: fff0

Explanation: -1 is internally represented as all 1’s. When left shifted four times the least significant 4 bits are filled with 0’s.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

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  1.  main()

{

char string[]=”Hello World”;

display(string);

}

void display(char *string)

{

printf(“%s”,string);

}

Answer:

Compiler Error: Type mismatch in redeclaration of function display

Explanation: In third line, when the function display is encountered, the compiler doesn’t know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

 

  1.  main()

{

int c=- -2;

printf(“c=%d”,c);

}

Answer:

c=2;

Explanation: Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.

Note: However you cannot give like –2. Because — operator can only be applied to variables as a decrement operator (eg., i–). 2 is a constant and not a variable.

 

  1.  #define int char

main()

{

int i=65;

printf(“sizeof(i)=%d”,sizeof(i));

}

Answer:

sizeof(i)=1

Explanation: Since the #define replaces the string int by the macro char

 

  1.  main()

{

int i=10;

i=!i>14;

Printf (“i=%d”,i);

}

Answer:

i=0

Explanation: In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

 

  1.  #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:

77

Explanation: p is pointing to character ‘\n’. str1 is pointing to character ‘a’ ++*p. “p is pointing to ‘\n’ and that is incremented by one.” the ASCII value of ‘\n’ is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98.

Now performing (11 + 98 – 32), we get 77(“M”); So we get the output 77 :: “M” (Ascii is 77).

 

  1.  #include<stdio.h>

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf(“%d—-%d”,*p,*q);

}

Answer:

SomeGarbageValue—1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays, but you are trying

to access the third 2D(which you are not declared) it will print garbage values.

*q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

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  1.  #include<stdio.h>

main()

{

struct xx

{

int x=3;

char name[]=”hello”;

};

struct xx *s;

printf(“%d”,s->x);

printf(“%s”,s->name);

}

Answer:

Compiler Error

Explanation: You should not initialize variables in declaration

 

  1.  #include<stdio.h>

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation: The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

 

  1.  main()

{

printf(“\nab”);

printf(“\bsi”);

printf(“\rha”);

}

Answer: hai

Explanation:

\n – newline

\b – backspace

\r – linefeed

20. main()

{

int i=5;

printf(“%d%d%d%d%d%d”,i++,i–,++i,–i,i);

}

Answer: 45545

Explanation: The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. And the evaluation is from right to left, hence the result.

 

  1.  #define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf(“%d”,i);

}

Answer: 64

Explanation: the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

 

  1.  main()

{

char *p=”hai friends”,*p1;

p1=p;

while(*p!=”) ++*p++;

printf(“%s %s”,p,p1);

}

Answer: ibj!gsjfoet

Explanation: ++*p++ will be parse in the given order

_ *p that is value at the location currently pointed by p will be taken

_ ++*p the retrieved value will be incremented

_ when; is encountered the location will be incremented that is p++ will be executed Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘’ and p1 points to p thus p1doesnot print anything.

 

  1. #include <stdio.h>

#define a 10

main()

{

#define a 50

printf(“%d”,a);

}

Answer: 50

Explanation: The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

 

  1.  #define clrscr() 100

main()

{

clrscr();

printf(“%d\n”,clrscr());

}

Answer: 100

Explanation: Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs. The input program to compiler looks like this :

main()

{

100;

printf(“%d\n”,100);

}

Note:

100; is an executable statement but with no action. So it doesn’t give any problem

 

  1.  main()

{

41printf(“%p”,main);

}8

Answer: Some address will be printed.

Explanation: Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

 

 

Linux Questions

When using useradd to create a new user account, which of the following tasks is not done automatically. Choose one: a. Assign a UID. b. Assign a default shell. c. Create the user’s home directory. d. Define the user’s home directory.

You issue the following command useradd -m bobm But the user cannot logon. What is the problem? Choose one: a. You need to assign a password to bobm’s account using the passwd command. b. You need to create bobm’s home directory and set the appropriate permissions. c. You need to edit the /etc/passwd file and assign a shell for bobm’s account. d. The username must be at least five characters long.

You have created special configuration files that you want copied to each user’s home directories when creating new user accounts. You copy the files to /etc/skel. Which of the following commands will make this happen? Choose one: a. useradd -m username b. useradd -mk username c. useradd -k username d. useradd -Dk username

Mary has recently gotten married and wants to change her username from mstone to mknight. Which of the following commands should you run to accomplish this? Choose one: a. usermod -l mknight mstone b. usermod -l mstone mknight c. usermod -u mknight mstone d. usermod -u mstone mknight

After bob leaves the company you issue the command userdel bob. Although his entry in the /etc/passwd file has been deleted, his home directory is still there. What command could you have used to make sure that his home directory was also deleted? Choose one: a. userdel -m bob b. userdel -u bob c. userdel -l bob d. userdel -r bob

All groups are defined in the /etc/group file. Each entry contains four fields in the following order. Choose one: a. groupname, password, GID, member list b. GID, groupname, password, member list c. groupname, GID, password, member list d. GID, member list, groupname, password

You need to create a new group called sales with Bob, Mary and Joe as members. Which of the following would accomplish this? Choose one: a. Add the following line to the /etc/group file: sales:44:bob,mary,joe b. Issue the command groupadd sales. c. Issue the command groupadd -a sales bob,mary,joe d. Add the following line to the /etc/group file: sales::44:bob,mary,joe

You attempt to use shadow passwords but are unsuccessful. What characteristic of the /etc/passwd file may cause this? Choose one: a. The login command is missing. b. The username is too long. c. The password field is blank. d. The password field is prefaced by an asterick.

You create a new user account by adding the following line to your /etc/passwd file. bobm:baddog:501:501:Bob Morris:/home/bobm:/bin/bash Bob calls you and tells you that he cannot logon. You verify that he is using the correct username and password. What is the problem? Choose one: a. The UID and GID cannot be identical. b. You cannot have spaces in the line unless they are surrounded with double quotes. c. You cannot directly enter the password; rather you have to use the passwd command to assign a password to the user. d. The username is too short, it must be at least six characters long.

Which of the following tasks is not necessary when creating a new user by editing the /etc/passwd file? Choose one: a. Create a link from the user’s home directory to the shell the user will use. b. Create the user’s home directory c. Use the passwd command to assign a password to the account. d. Add the user to the specified group.

You create a new user by adding the following line to the /etc/passwd file bobm::501:501:Bob Morris:/home/bobm:/bin/bash You then create the user’s home directory and use the passwd command to set his password. However, the user calls you and says that he cannot log on. What is the problem? Choose one: a. The user did not change his password. b. bobm does not have permission to /home/bobm. c. The user did not type his username in all caps. d. You cannot leave the password field blank when creating a new user.

 

 

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