Our website is made possible by displaying online advertisements to our visitors.Please consider supporting us by disabling your ad blocker.
Drop Down MenusCSS Drop Down MenuPure CSS Dropdown Menu
Drop Down MenusCSS Drop Down MenuPure CSS Dropdown Menu

Translate it in your own Language

Print this Job Post

Print Friendly and PDF

Tuesday, June 10, 2014

51. What is Mail Gateway?
Answer:
It is a system that performs a protocol translation between different electronic mail delivery protocols.

52.What is IGP (Interior Gateway Protocol)?
Answer:
It is any routing protocol used within an autonomous system.

53.What is EGP (Exterior Gateway Protocol)?
Answer:
It is the protocol the routers in neighboring autonomous systems use to identify the set of networks that can be reached within or via each autonomous system.

54.What is autonomous system?
Answer:
It is a collection of routers under the control of a single administrative authority and that uses a common Interior Gateway Protocol.

55.What is BGP (Border Gateway Protocol)?
Answer:
It is a protocol used to advertise the set of networks that can be reached with in an autonomous system. BGP enables this information to be shared with the autonomous system. This is newer than EGP (Exterior Gateway Protocol).

56.What is Gateway-to-Gateway protocol?
Answer:
It is a protocol formerly used to exchange routing information between Internet core routers.

57.What is NVT (Network Virtual Terminal)?
Answer:
It is a set of rules defining a very simple virtual terminal interaction. The NVT is used in the start of a Telnet session.

58.What is a Multi-homed Host?
Answer:
It is a host that has a multiple network interfaces and that requires multiple IP addresses is called as a Multi-homed Host.

59.What is Kerberos?
Answer:
It is an authentication service developed at the Massachusetts Institute of Technology. Kerberos uses encryption to prevent intruders from discovering passwords and gaining unauthorized access to files.

60.What is OSPF?
Answer:
It is an Internet routing protocol that scales well, can route traffic along multiple paths, and uses knowledge of an Internet's topology to make accurate routing decisions.

61.What is Proxy ARP?
Answer:
It is using a router to answer ARP requests. This will be done when the originating host believes that a destination is local, when in fact is lies beyond router.

62.What is SLIP (Serial Line Interface Protocol)?
Answer:
It is a very simple protocol used for transmission of IP datagrams across a serial line.

63.What is RIP (Routing Information Protocol)?
Answer:
It is a simple protocol used to exchange information between the routers. 

64.What is source route?
Answer:
It is a sequence of IP addresses identifying the route a datagram must follow. A source route may optionally be included in an IP datagram header. 

                        « Prev 123 Next »

26.What is ICMP?
Answer:
ICMP is Internet Control Message Protocol, a network layer protocol of the TCP/IP suite used by hosts and gateways to send notification of datagram problems back to the sender. It uses the echo test / reply to test whether a destination is reachable and responding. It also handles both control and error messages.

27.What are the data units at different layers of the TCP / IP protocol suite?
Answer:
The data unit created at the application layer is called a message, at the transport layer the data unit created is called either a segment or an user datagram, at the network layer the data unit created is called the datagram, at the data link layer the datagram is encapsulated in to a frame and finally transmitted as signals along the transmission media.

28.What is difference between ARP and RARP?
Answer:
The address resolution protocol (ARP) is used to associate the 32 bit IP address with the 48 bit physical address, used by a host or a router to find the physical address of another host on its network by sending a ARP query packet that includes the IP address of the receiver.
The reverse address resolution protocol (RARP) allows a host to discover its Internet address when it knows only its physical address.

29.What is the minimum and maximum length of the header in the TCP segment and IP datagram?
Answer:
The header should have a minimum length of 20 bytes and can have a maximum length of 60 bytes.   
                     
30.What is the range of addresses in the classes of internet addresses? 
Answer:
Class A      0.0.0.0      - 127.255.255.255
Class B 128.0.0.0 - 191.255.255.255
Class C 192.0.0.0 - 223.255.255.255
Class D 224.0.0.0 - 239.255.255.255
Class E 240.0.0.0 - 247.255.255.255

31.What is the difference between TFTP and FTP application layer protocols?
Answer:
The Trivial File Transfer Protocol (TFTP) allows a local host to obtain files from a remote host but does not provide reliability or security. It uses the fundamental packet delivery services offered by UDP.
The File Transfer Protocol (FTP) is the standard mechanism provided by TCP / IP for copying a file from one host to another. It uses the services offer by TCP and so is reliable and secure. It establishes two connections (virtual circuits) between the hosts, one for data transfer and another for control information.

32.What are major types of networks and explain?
Answer:

Server-based network:

Peer-to-peer network, computers can act as both servers sharing resources and as clients using the resources.

Peer-to-peer network

Server-based networks provide centralized control of network resources and rely on server computers to provide security and network administration

33.What are the important topologies for networks?
Answer:

BUS topology:

In this each computer is directly connected to primary network cable in a single line.
Advantages:
Inexpensive, easy to install, simple to understand, easy to extend.

STAR topology:

In this all computers are connected using a central hub.
Advantages:
Can be inexpensive, easy to install and reconfigure and easy to trouble shoot physical problems.

RING topology:

In this all computers are connected in loop.
Advantages:
All computers have equal access to network media, installation can be simple, and signal does not degrade as much as in other topologies because each computer regenerates it.

34.What is mesh network?
Answer:
A network in which there are multiple network links between computers to provide multiple paths for data to travel.

35.What is difference between baseband and broadband transmission?
Answer:
In a baseband transmission, the entire bandwidth of the cable is consumed by a single signal. In broadband transmission, signals are sent on multiple frequencies, allowing multiple signals to be sent simultaneously.

36.Explain 5-4-3 rule?
Answer:
In a Ethernet network, between any two points on the network ,there can be no more than five network segments or four repeaters, and of those five segments only three of segments can be populated.

37.What MAU?
Answer:
In token Ring , hub is called Multistation Access Unit(MAU).

38.What is the difference between routable and non- routable protocols?
Answer:
Routable protocols can work with a router and can be used to build large networks. Non-Routable protocols are designed to work on small, local networks and cannot be used with a router

39.Why should you care about the OSI Reference Model?
Answer:
It provides a framework for discussing network operations and design.

40. What is logical link control?
Answer:
One of two sublayers of the data link layer of OSI reference model, as defined by the IEEE 802 standard. This sublayer is responsible for maintaining the link between computers when they are sending data across the physical network connection. 

41. What is virtual channel?
Answer:
Virtual channel is normally a connection from one source to one destination, although multicast connections are also permitted. The other name for virtual channel is virtual circuit.

42.What is virtual path?
Answer:
Along any transmission path from a given source to a given destination, a group of virtual circuits can be grouped together into what is called path.

43. What is packet filter?
Answer:
Packet filter is a standard router equipped with some extra functionality. The extra functionality allows every incoming or outgoing packet to be inspected. Packets meeting some criterion are forwarded normally. Those that fail the test are dropped.

44.What is traffic shaping?
Answer:
One of the main causes of congestion is that traffic is often busy. If hosts could be made to transmit at a uniform rate, congestion would be less common. Another open loop method to help manage congestion is forcing the packet to be transmitted at a more predictable rate. This is called traffic shaping.

45.What is multicast routing?
Answer:
Sending a message to a group is called multicasting, and its routing algorithm is called multicast routing.

46.What is region?
Answer:
When hierarchical routing is used, the routers are divided into what we will call regions, with each router knowing all the details about how to route packets to destinations within its own region, but knowing nothing about the internal structure of other regions.

47.What is silly window syndrome? 
Answer:
It is a problem that can ruin TCP performance. This problem occurs when data are passed to the sending TCP entity in large blocks, but an interactive application on the receiving side reads 1 byte at a time.

48.What are Digrams and Trigrams?
Answer:
The most common two letter combinations are called as digrams. e.g. th, in, er, re and an. The most common three letter combinations are called as trigrams. e.g. the, ing, and, and ion.

49.Expand IDEA.
Answer:
IDEA stands for International Data Encryption Algorithm.

50.What is wide-mouth frog?
Answer:
Wide-mouth frog is the simplest known key distribution center (KDC) authentication protocol.

                         « Prev 123 Next »

Note :

All the programs are tested under Turbo C++ 3.0, 4.5 and Microsoft VC++ 6.0 compilers. 
It is assumed that,

  • Programs run under Windows environment,
  • The underlying machine is an x86 based system,
  • Program is compiled using Turbo C/C++ compiler.
  • The program output may depend on the information based on this assumptions (for example sizeof(int) == 2 may be assumed). 


1) 
class Sample
{
public:
        int *ptr;
        Sample(int i)
        {
        ptr = new int(i);
        }
        ~Sample()
        {
        delete ptr;
        }
        void PrintVal()
        {
        cout << "The value is " << *ptr;
        }
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
Say i am in someFunc 
Null pointer assignment(Run-time error)
Explanation:
As the object is passed by value to SomeFunc  the destructor of the object is called when the control returns from the function. So when PrintVal is called it meets up with ptr  that has been freed.The solution is to pass the Sample object  by reference to SomeFunc:
void SomeFunc(Sample &x)
{
cout << "Say i am in someFunc " << endl;
}
because when we pass objects by refernece that object is not destroyed. while returning from the function.

2)
Which is the parameter that is added to every non-static member function when it is called?
Answer:
‘this’ pointer

3) 
class base
{
  public:
  int bval;
  base(){ bval=0;}
 };
class deri:public base
 {
   public:
   int dval;
   deri()
    { 
     dval=1;
   }
};
void SomeFunc(base *arr,int size)
{
for(int i=0; i<size; i++,arr++)
cout<<arr->bval;
cout<<endl;
}
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
}
Answer:
 00000
 01010
Explanation:  
The function SomeFunc expects two arguments.The first one is a pointer to an array of base class objects and the second one is the sizeof the array.The first call of someFunc calls it with an array of bae objects, so it works correctly and prints the bval of all the objects. When Somefunc is called the second time the argument passed is the pointeer to an array of derived class objects and not the array of base class objects. But that is what the function expects to be sent. So the derived class pointer is promoted to base class pointer and the address is sent to the function. SomeFunc() knows nothing about this and just treats the pointer as an array of base class objects. So when arr++ is met, the size of base class object is taken into consideration and is incremented by sizeof(int) bytes for bval (the deri class objects have bval and dval as members and so is of size >= sizeof(int)+sizeof(int) ). 

4) 
class base
{
public:
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
 baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from base
Explanation:
As we have seen in the previous case, SomeFunc expects a pointer to a base class. Since a pointer to a derived class object is passed, it treats the argument only as a base class pointer and the corresponding base function is called. 

5) 
class base
{
 public:
 virtual void baseFun(){ cout<<"from base"<<endl;}
};
 class deri:public base
{
 public:
 void baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
 baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from derived 
Explanation:
Remember that baseFunc is a virtual function. That means that it supports run-time polymorphism. So the function corresponding to the derived class object is called. 

6).
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
Answer : 
Compiler Error: 'ra',reference must be initialized
Explanation : 
Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned,whereas references can only be initialized. So this code issues an error.

7).
const int size = 5;
void print(int *ptr)
{
cout<<ptr[0];
}
void print(int ptr[size])
{
cout<<ptr[0];
}
void main()
{
int a[size] = {1,2,3,4,5};
int *b = new int(size);
print(a);
print(b);
}
Answer:
Compiler Error : function 'void print(int *)' already has a body
Explanation:
Arrays cannot be passed to functions, only pointers (for arrays, base addresses) can be passed. So the arguments int *ptr and int prt[size] have no difference  as function arguments. In other words, both the functoins have the same signature and so cannot be overloaded. 

9).
class some
{
public:
~some()
{
cout<<"some's destructor"<<endl;
}
};
void main()
{
some s;
s.~some();
}
Answer:
some's destructor
some's destructor
Explanation:
Destructors can be called explicitly. Here 's.~some()' explicitly calls the destructor of 's'. When main() returns, destructor of s is called again,hence the result.

11).
#include <iostream.h>
class fig2d
{
int dim1;
int dim2;
public:
fig2d() { dim1=5; dim2=6;}
virtual void operator<<(ostream & rhs);
};
void fig2d::operator<<(ostream &rhs)
{
rhs <<this->dim1<<" "<<this->dim2<<" ";
}
/*class fig3d : public fig2d
{
int dim3;
public:
fig3d() { dim3=7;}
virtual void operator<<(ostream &rhs);
};
void fig3d::operator<<(ostream &rhs)
{
fig2d::operator <<(rhs);
rhs<<this->dim3;
}
*/
void main()
{
fig2d obj1;
//fig3d obj2;
obj1 << cout;
//obj2 << cout;
}
Answer : 
5 6 
Explanation:
In this program, the << operator is overloaded with ostream as argument. This enables the 'cout' to be present at the right-hand-side. Normally, 'cout' is implemented as global function, but it doesn't mean that 'cout' is not possible to be overloaded as member function.
Overloading << as virtual member function becomes handy when the class in which it is overloaded is inherited, and this becomes available to be overrided. This is as opposed to global friend functions, where friend's are not inherited.

12).
class opOverload
{
public:
bool operator==(opOverload temp);
};
bool opOverload::operator==(opOverload temp)
{
if(*this  == temp )
{
cout<<"The both are same objects\n";
return true;
}
else
{
cout<<"The both are different\n";
return false;
}
}
void main()
{
opOverload a1, a2;
a1= =a2;
}
Answer : 
Runtime Error: Stack Overflow
Explanation :
Just like normal functions, operator functions can be called recursively. This program just illustrates that point, by calling the operator == function recursively, leading to an infinite loop. 

13).
class complex
{
double re;
double im;
public:
complex() : re(1),im(0.5) {}
bool operator==(complex &rhs);
operator int(){}
};
bool complex::operator == (complex &rhs){
if((this->re == rhs.re) && (this->im == rhs.im))
return true;
else
return false;
}
int main()
{
complex  c1;
cout<<  c1;
}
Answer : 
Garbage value
Explanation:
The programmer wishes to print the complex object using output
re-direction operator,which he has not defined for his lass.But the compiler instead of giving an error sees the conversion function and converts the user defined object to standard object and prints some garbage value.

14).
class complex
{
double re;
double im;
public:
complex() : re(0),im(0) {}
complex(double n) { re=n,im=n;};
complex(int m,int n) { re=m,im=n;}
void print() { cout<<re; cout<<im;}
};
void main()
{
complex c3;
double i=5;
c3 = i;
c3.print();
}
Answer: 
5,5 
Explanation:
Though no operator= function taking complex, double is defined, the double on the rhs is converted into a temporary object using the single argument constructor taking double and assigned to the lvalue.

15).
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
Answer : 
Compiler Error: 'ra',reference must be initialized
Explanation : 
Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned,whereas references can only be initialized. So this code issues an error.

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

51) 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’.

52) 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’.

53) main()
       {
        int  i, n;
        char *x = “girl”;
        n = strlen(x);
        *x = x[n];
        for(i=0; i<n; ++i)
       {
       printf(“%s\n”,x);
       x++;
      }
     }  
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”.The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position.  Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

54) int i,j;
       for(i=0;i<=10;i++)
       {
        j+=5;
       assert(i<5);
      }
Answer: 
Runtime error: Abnormal program termination. 
assert failed (i<5), <file name>,<line number> 
Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,#undef NDEBUG and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of.  
  
55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}
Answer:
 i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :

  • a: The SEEK_SET sets the file position marker to the starting of the file.
  • b: The SEEK_CUR sets the file position marker to the current position of the file.


58) main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute? 
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it  reads all character upto another quotation mark.

59) What is the problem with the following code segment? while ((fgets(receiving array,50,file_ptr)) != EOF);
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

61) main()
{
char *cptr,c;
void *vptr,v;
c=10;  v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

63) main()
{
char not;
not=!2;
printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

64) #define FALSE -1
#define TRUE   1
#define NULL   0
main() 
                     {
  if(NULL)
  puts("NULL");
  else if(FALSE)
  puts("TRUE");
  else
  puts("FALSE");
 }
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main()
{
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
    printf("%d is a leap year");
else
    printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.

67)   #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb: 

#defines are used for textual replacement whereas typedefs are used for declaring new types.

68) int i=10;
main()
{
  extern int i;
           {
    int i=20;
     {
      const volatile unsigned i=30;
      printf("%d",i);
     }
     printf("%d",i);
  }
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

69) main()
       {
        int *j;
        {
         int i=10;
         j=&i;
         }
         printf("%d",*j);
         }
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

71) #include<stdio.h>
        main()
        {
         const int i=4;
         float j;
         j = ++i;
         printf("%d  %f", i,++j);
        }
Answer:
Compiler error 
Explanation:
i is a constant. you cannot change the value of constant 

72) #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:
garbagevalue..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 meAnswer:it will print first element of 3D array.

73) #include<stdio.h>
       main()
       {
        register i=5;
        char j[]= "hello";                     
        printf("%s  %d",j,i);
       }
Answer:
hello 5
Explanation:
if you declare i as register  compiler will treat it as ordinary integer and it will take integer value. i value may be  stored  either in register  or in memory.

74) main()
       {
         int i=5,j=6,z;
         printf("%d",i+++j);
       }
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)    

                    « Prev 1234 Next »

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

76) struct aaa
       {
        struct aaa *prev;
        int i;
        struct aaa *next;
       };
      main()
      {
       struct aaa abc,def,ghi,jkl;
       int x=100;
       abc.i=0;abc.prev=&jkl;
       abc.next=&def;
       def.i=1;def.prev=&abc;def.next=&ghi;
       ghi.i=2;ghi.prev=&def;
       ghi.next=&jkl;
       jkl.i=3;
       jkl.prev=&ghi;
       jkl.next=&abc;
       x=abc.next->next->prev->next->i;
       printf("%d",x);
      }
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i 
this one points to "ghi" node the value of at particular node is 2.

77)
 struct point
 {
 int x;
 int y;
 };
struct point origin,*pp;
main()
{
pp=&origin;
printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
printf("origin is (%d%d)\n",pp->x,pp->y);

Answer:
origin is(0,0)
origin is(0,0) 
Explanation:
pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator. 

Note: 

Since structure point  is globally declared x & y are initialized as zeroes 

78)
 main()
{
 int i=_l_abc(10);
  printf("%d\n",--i);
}
int _l_abc(int i)
{
 return(i++);
}
Answer:
9
Explanation: 
return(i++) it will first return i and then increments. i.e. 10 will be returned.

79)
 main()
{
 char *p;
 int *q;
 long *r;
 p=q=r=0;
 p++;
 q++;
 r++;
 printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator  when applied to pointers increments address according to their corresponding data-types.

 80)
 main()
{
 char c=' ',x,convert(z);
 getc(c);
 if((c>='a') && (c<='z'))
 x=convert(c);
 printf("%c",x);
}
convert(z)
{
  return z-32;
}
Answer: 
Compiler error
Explanation:
declaration of convert and format of getc() are wrong. 

81)
 main(int argc, char **argv)
{
 printf("enter the character");
 getchar();
 sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
 return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.  

82)
# include <stdio.h>
int one_d[]={1,2,3};
main()
{
 int *ptr; 
 ptr=one_d;
 ptr+=3;
 printf("%d",*ptr);
}
Answer:
garbage value
Explanation:
ptr pointer is pointing to out of the array range of one_d.

83)
# include<stdio.h>
aaa() 
{
  printf("hi");
 }
bbb()
{
 printf("hello");
 }
ccc()
{
 printf("bye");
 }
main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}
Answer:
bye 
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

85)
#include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop  
Explanation:
The condition is checked against EOF, it should be checked against NULL.

86)
main()
{
 int i =0;j=0;
 if(i && j++)
 printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0 
Explanation:
The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed.  The values of i and j remain unchanged and get printed.
      
87)
 main()
{
 int i;
 i = abc();
 printf("%d",i);
}
abc()
{
 _AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000. 

88)
int i;
main()
{
int t;
for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
 4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution will be,
           t        i        x
          4       0      -4
          3       1      -2
          2       2       0
          
89)
main()
{
  int a= 0;
  int b = 20;
  char x =1;
  char y =10;
  if(a,b,x,y)
  printf("hello");
 }
Answer:
hello 
Explanation:
The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

90)
main()
{
 unsigned int i;
 for(i=1;i>-2;i--)
 printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop. 

91) 
In the following pgm add a  stmt in the function  fun such that the address of  'a' gets stored in 'j'.
main()
{
  int * j;
  void fun(int **);
  fun(&j);
 }
 void fun(int **k) {
  int a =0;
  /* add a stmt here*/
 }
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
      
92) 
What are the following notations of defining functions known as?
i.      int abc(int a,float b)
        {
          /* some code */
        }
ii.    int abc(a,b)
       int a; float b;
        {
          /* some code*/
        }
Answer:
i.  ANSI C notation
ii. Kernighan & Ritche notation 

93)
main()
{
char *p;
p="%d\n";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

94)
main()
{
 char a[100];
 a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
 abc(a);
}
abc(char a[])
{
 a++; 
 printf("%c",*a);
 a++;
 printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.
                
95)
func(a,b)
int a,b;
{
 return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !\n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
 }
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function  2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

96)
void main()
{
static int i=5;
if(--i)
{
main();
printf("%d ",i);
}
}
Answer:
 0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

97)
void main()
{
int k=ret(sizeof(float));
printf("\n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
return(ret);
}
Answer:
 Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed,  after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

98)
void main()
{
 char a[]="12345\0";
 int i=strlen(a);
 printf("here in 3 %d\n",++i);
}
Answer: 
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre increment in the printf statement, the 6 will be printed.

99)
void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u \n",gotit=--giveit);
}
Answer:
 0 65535

100)
void main()
{
int i;
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 null character returns 1 which makes the if statement true, thus "Ok here" is printed

                     « Prev 1234 Next »

More Details About CrackMNC

Follow Our Google+ Page

Copyright @ CrackMNC 2014-2024
Divas Nikhra Theme by Crack MNC