hcl technologies paper & interview –
Bhaskar Reddy here.. I was attended for HCL Technologies in Anna University on Nov 26th and I got placed. Here I am presenting my experience in written test and interview. I hope it may be useful for you..
Written test consists of 4 sections. In each section we have to get minimum marks. Which means it’s a sectional cut off. Also negative marking is there. They will deduct ¼ of each wrong answer… Ok… You just answer all questions with lot of confidence. They will give all 4 sections question paper at a time.. so you can start any section.
Total paper consists of 65 Questions and provided 90 min time. So that you can answer each and every section very carefully.. ok
SECTION-I :(Microprocessor (8086 only) + Networks + OS + Fundamentals of computers)
This section consists of 15 Questions. Here I remember some questions….
1) Non Maskable Interrupt is a….
a) NMI b) Software interrupts
c) Hardware interrupts d) Software and hardware interrupt
Ans: b
2) Which of the following is error correction and deduction?
a) Hamming code b) CRC c) VRC d) None
Ans: b
3) When you switch on your computer, which of the following component affect first?
a) Mother board b) SMPS c) Peripherals d) None
Ans : a
4) Which of the following function transports the data?
a) TCP/IP b) Transport layer c) TCP d) None
Ans: c
5) Which of the following does not consists address?
a) IP b) TCP/IP c) Network d) Transport
Ans:a
6) They given like this….. And some conditions?
a) pre order b) post order c) In order d) None
Ans:c
7) Authentication means…
a) Verifying authority b) Source c) Destination d) Request
Ans: a
8) Symorphous is used for
a) Analysis b) Synchournization c) Asynchrouns d) None
Ans: b
9) There are five nodes. With that how many trees we can make?
a) 27 b) 28 c) 30 d) 29
Ans: c (Check the ans not sure)
10) Traverse the given tree using in order, Preorder and Post order traversals.
Given tree:
Ø Inorder : D H B E A F C I G J
Ø Preorder: A B D H E C F G I J
Ø Postorder: H D E B F I J G C A
And some more questions…. I dint remember those questions
SECTION -II (C language Basics and programming) this section consists of 20 Questions…… All are programs only..
1. main()
{
printf(“%x”,-1<<4);
}
a) fff0 b) fffb c) ff0 d) none
Ans: a
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.
2. main()
{
char *p;
p=”Hello”;
printf(“%c\n”,*&*p);
}
a) e b) H c) some address d) ome garbage value
Ans: b
Explanation: * is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here
p points to the first character in the string “Hello”. *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.
3. void main()
{
int i=5;
printf(“%d”,i++ + ++i);
}
a) 11 b) 12 c) 10 d) output cant be predicted
Ans: d
Explanation: Side effects are involved in the evaluation of i
4. main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
}
a) 100, 100, 100, 2 b) 101,101,101,2 c) 114,104,102,3 d) none
Ans: c
Explanation:The given array is a 3-D one. It can also be viewed as a 1-D array.
2
4
7
8
3
4
2
2
2
3
3
4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element . since the indirection ***a gives the value. Hence, the first line of the output. for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output is C is correct answer..
5. main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
a) 111
222
332
443
b) 111
222
333
344
c) 111
222
333
444
d) None
Ans: b
6. #include
main()
{
const int i=4;
float j;
j = ++i;
printf(“%d %f”, i,++j);
}
a) 8 b) 5 c) compile error d) syntax error
Ans: c
7. main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf(“%p…%p…%p”,p,q,r);
}
a) 001.. 100..0002 b) 0001…0002…0004
c) 001.. 002..004 d) none
ans: b
8. main()
{
unsigned int i;
for(i=1;i>-2;i–)
printf(“HCL Technologies”);
}
a) HCL b) Technologies c) HCL Technologies d) None
Ans: None(Plz Check the answer)
9. main()
{
int a[10];
printf(“%d”,*a+1-*a+3);
}
a) 4 b) 5 c) 6 d) None
Ans : a
10. main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf(“%d\n”,++k);
printf(“%f\n”,f<<2);
printf(“%lf\n”,f%g);
printf(“%lf\n”,fmod(f,g));
}
a) Line no 5: Error: Lvalue required
b) Line no 5: Error: Link error
c) Compile error d) None
Ans: a
11. int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
{
int x=10,y=20;
swap(&x,&y);
printf(“x= %d y = %d\n”,x,y)
}
a) x=10 y=20 b) x=20 y=10 c) x=30 y=20 d) none
Ans: b
12. main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf(“%d”,i);
}
a) 665 b) 565 c) 556 d) none
Ans: c
13.main()
{
float me=1.1;
double you=1.1;
if(me==you)
printf(“IloveU”);
else
printf(“I Hate U”)
}
a) I love u b) I hate u c) floating point error d) Compile error
Ans: b
14.
enum colors {BLACK,BLUE,GREEN}
main()
{
printf(“%d..%d..%d”,BLACK,BLUE,GREEN);
return(1);
}
a) 1..2..3 b) 0..1..2 c) 2..3..4 d) none
Ans: b
Some more questions given.. I dint remember… Be prepare all basic concept in C.. so that you can answer very easily..
SECTION -III (Data structures and C++) this section consists of 10 Questions.. Each question carry 2 marks.. so they deduct ½ mark for wrong answer….
1) #include
main()
{
char s[]={‘a’,’b’,’c’,’\n’,’c’,’\0′};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf(“%d”,++*p + ++*str1-32);
}
a)97 b) M c)76 d) none
Ans: b
2) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
a) address of array b) Compile error c) Lvalue required d) none
Ans: c
3) 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);
}
a) 3 b) 2 c) 4 d) 5
Ans: b
4) main(int argc, char **argv)
{
printf(“enter the character”);
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
a) compile error b) L value required c) Syntax error d) None
Ans: a
5)
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();
}
a) say I am in someFunc b) say I am in someFunc and runtime error
c) say I am in someFunc and null value d) none
ans: b
and some questions given in DATASTURES.. those are based on linked lists only also very big programs.. so you have to do it very careful
SECTION-IV (General aptitude+1 passage+Logical) this section consists of 20 questions.. Each question carry 1 mark..
1) Sandhya and Bhagya is having the total amount of 12000. In that amount bhagya has deducted 3600 as less as sandhya. So what is their shared amount?
a) 2800 b) 3600 c) 4800 d) 9600
Ans : c
2) Six persons have to present at certain meeting. The conditions are,
A present P should present
M present T should present
K present P should present
If A and P present I should be there in meeting
If M and T present D should be absent
If K and P present P should present
Based on this they given some questions.. Those are easy only.. you can do it easily..
3) Here they given passage following some questions.. Poetry explaining her experience with HINDI latest songs.. and comparing with old songs.. also she is a good singer. Like this they given a big passage.. so read it carefully at a time.. so that u can save your time
4) Dhana and Lavanya started running at same point.. But dhana started in anti clock wise direction, Lavanya started in clockwise direction.. Dhana met lavanya at 900 m . where as lavanya met dhana at 800 m.. so how much distance they covered each other?
a) 1700 b) 900 c) 1800 d) data is insufficient
Ans: d
5) This question is base on Arithmetic mean like algebra.. a n+2 = (7+an)/5… Initially a0= 0…. so what is the value of a2?
a) 5/2 b) 7/2 c) 7/5 d) none
Ans : c
6) Here they given two statements, based on that they gave some questions
Statement I : I is enough to answer
Statement II: I and II is enough to answer
i) Raja can do a piece of work in 9 days.. and govardhan can do a piece of work in 8 days. In how many days they will complete the work alternatively..
Statement I: They both do in 72/17 days
Statement II : A alone can do 1/9 days
a) I b) I and II c) II d) none
Like this they given some questions….
My Technical interview held like this..
Intwr: Hi, I am Mahesh butt
Me: hi sir I am bhaskarreddy
Intwr: ok.. Which project u did in your curriculum?
Me: sir my project is steganography
Intwr: oh… What is steganography?Why you choose this project? What’s your project purpose?
What is encryption? How you embedded the selection of sector? Which code u used?Which algorithm you used? Can you explain me?
Mostly I answered all questions.. Also he gave me paper and pen… Like that I explained to him very clearly with diagrams.
Intwr: So bhaskar u knows C well? You are good at C language? (Because I wrote C and .NET in my resume)
Me: no… Sir, I am not proficient in C, but logically I am good and I am in intermediate stage.
Intwr: oh..ok.. What is your area of interest?
Me: Digital Electronics.. sir
Intwr: Are you sure?
Me: yes sir..(With lot of confidence)
Intwr:
A0
B0
\
c=1
A1
B1
Bhaskar you use your entire Digital electronics, but I Want inside circuit which always shows the output as 1.
The conditions are,
When x=0: A when condition is true , output is 1
When x=1: A>B When condition is false , output is 0
Consider A=A0A1
Consider B= B0B1
Here he given the time of 15 mini to solve above problem… But I did it with in 5 mini with logic gates.
Then he asked me to explain the entire circuit?
I explained to him…….
Ok.. Bhaskar I am giving one more 15 mini time to solve above question… But this time you have to do with Multiplexers…
Yes.. here I designed the circuit and I explained to him… Here I impressed him a lot..
Intwr: Ok.. Good… Draw the block diagram of Feed back amplifier?
Me: I drawn the diagram
Intwr: You just add forward and reverse bias diodes to the feedback? I did it.. ok.. will this circuit runs
Me: Yes… Sir
Intwr: why
Me: No sir
Intwr: why
Me: Bcz some nose occurs in output and said some thing..
Intwr: What is pointer?
What is data sturucture?
What is stack?Queue?
Explain the Doubly linked list program?
I explained to him very clearly..
Intwr:
main()
{
int *j;
{
int i=10;
j=&i;
}
printf(“%d”,*j);
printf(“%d”, j);
}
What is the output?
Me: I did it correctly
Intrwr: Write a program for Fibonacci series? But I want to print only last value?
Me: I wrote program for Fibonacci series.. But I don’t know how to print last value..
Intwr: are you shivering
Me: yes… Sir…
Intwr: why bhaskar.. You are good at your subject why cant you do this???
Me: I tried it.. at last I got it
Intwr: Ok bhaskar.. Now you can go…..
Me: Sir… Am I fit for HCL?
Intwr: Bhaskar lets see your dreams comes true
Me: Thank you sir..
After some time they called me to HR round.. That time only I decided that I cleared technical round..
HR: Hi… I am priya
Me: I am Bhaskar Reddy
HR: Tell me some thing about you?
Why you joined in Bhajarang?
In which domain you will work?
And some more simple questions?
So after that she gave me offer letter.. so that was the great exciting moment in my life..
So friends don’t ever loose your hopes.. Keep on trying.. Surely one good day is waiting for you.. Till then you do Hard work.. ok.. Hard work never fails.. GOOD LUCK
By, BHASKAR REDDY M
