There are four steps
1. Preprocessing
2. Compilation
3. Assembling
4. Linker
| STEPS | INPUT | OUTPUT | GCC |
| Pre Processing | source file | preprocessed output (removing all # defines & replacing macros) | gcc -E a.c |
| Compilation | Preprocessed output file | Assembly language code | gcc -S a.c |
| Assembling | Assembly code | Machine code | gcc -c a.c |
| Linker | Machine code | executable file | gcc a.c |
Had 5 rounds of interview
1. Telephonic technical Round
* steps of compilation
* storage classes & memory allocation
* pass by value & pass by reference
* questions on pointers
* stack & heap in C
* Reverse a linked list
* Middle element of a linked list
* difference between insertion & quick sort
* time complexities of all sorting algorithms
* difference between TCP & UDP
2. Technical round-1 (F2F)
* TCP/IP layer
* how ping works
* IP header
* questions about data link layer
3. Technical round-2 (F2F)
* Find a duplicate in an array of 1 to n elements
* swap without comparison
* deletion of a linked list
* find a loop in a linked list
* set a bit in a integer
* stack & heap - malloc & calloc
* some questions in pointers
4. Technical round-3 (F2F)
* find the sum of array using recursion
* pattern matching
* find the intersection of two linked list and questions on projects done.
5. Manager round (F2F)
* how trace route works
* how route works
* ipconfig works
* some HR questions like , how you manage pressure, strength etc.,
Given two sorted arrays A & B where size of A is m & size of B is m+n . B contains only n number of elements & m empty positions. Sorting should not take any extra memory.
#include<stdio.h>
main()
{
int A[]={1,3,7,10};
int B[7]={2,4,11};
int size_A;
int size_B;
int i;
size_A=sizeof(A)/sizeof(int);
size_B=sizeof(B)/sizeof(int);
size_A--;
size_B--;
int s=size_B;
int B_Elements= (size_B - size_A);
B_Elements--;
while((size_A >=0) && (B_Elements >=0))
{
if( (A[size_A] ) >= (B[B_Elements])){
B[size_B]=A[size_A];
printf("\n%d > = %d",A[size_A ] , B[B_Elements ]);
size_A--;
size_B--;
}
else
{
printf("\n%d < = %d",A[size_A ] , B[B_Elements ]);
B[size_B] = B[B_Elements];
size_B--;
B_Elements--;
}
}
if ( size_A >=0 ){
while ( size_A >= 0 ){
B[size_B] = A[size_A];
size_A--;
size_B--;
}
}
else if ( B_Elements >=0 ) {
while (B_Elements >=0){
B[size_B] = B[size_B];
B_Elements--;
size_B--;
}
}
else {
printf("Merging done");
}
printf("\n");
for(i=0;i<=s;i++)
printf("%d ",B[i]);
}
public class kadane {
public static void main(String args[])
{
System.out.println("Kadane Algorithm");
int A[]={-2, -3, 4, -1, -2, 1, 5, -3};
int max , max_final , i;
max=max_final=0;
int size=A.length;
for(i=0;i<size;i++)
{
max=max+A[i];
if(max < 0)
max=0;
if(max_final<max)
max_final=max;
}
System.out.println("Maximum sum : "+max_final);
}
}
Output:
Maximum sum : 7
1. Given a binary tree & a sum value. Find the two nodes in a binary tree whose sum is equal to the given sum.
2. Print all the permutations of a string.
3. Detect and remove the loop in linked list.
4. Given a binary tree , find the maximum rooted subtree (whose sum of its child is greater than other sub trees).
5. Difference between abstract class & packages in Java.
6. Convert Binary Search Tree to Sorted Doubly linked list.
7. Maximum circular subarry sum.
8. Given 3 numbers. Find the lease valid date formed by those numbers.
eg: 1,2 ,3. Output should be : 3/2/1. Note: should take care of cases like valid month & date & leap year.
9. Given binary tree . Print the sum of all the leaf nodes.
10. Given two sorted arrays , ‘A’ of size m and ‘B’ of size n+m. Array B will contain numbers in n positions and ‘m’ empty positions. Merget two Arrays. Note: Should not use extra memory.
6. Give sizeof(a) in bytes for the following: (assume 32bit PC)
char *a[10];
char (*a)[10];
char (**a)[20];
char (*a)(int x, char c);
Ans: 40 , 4 , 4 , 4
7. Find out the output :
#include<stdio.h>
main()
{
int n1 = 10, n2 = 12;
if( n2=!0)
printf("Result is:%d\n",(n1/n2));
else
printf("cannot divide by zero\n");
}
Ans: Result is:10
8. Find out the output:
#include<stdio.h>
main()
{
long int one_million;
one_million=1,000,000;
printf("One million is %d",one_million);
}
Ans: One million is 1
9. what is the output ?
#include<stdio.h>
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
main()
{
printf("%s %s\n",h(f(1,2)),g(f(1,2)));
}
Ans: 12 f(1,2)
10. what is the output?
#include<stdio.h>
main()
{
static int j=printf("%d",456);;
printf("%d",j);
}
Ans: Error
11. What is the output ?
#include<stdio.h>
main()
{
int i = 5;
int j;
j=sizeof(i++);
printf("%d",i);
}
Ans: 5
12. What is the output of z?
#include<stdio.h>
main()
{
int z,x=5,y=-10,a=4,b=2;
z=x++ - --y * b / a;
}
Ans: 10
1. what is the output of the following program ? “alloc failed or alloc successful” ?
#include<stdio.h>
void alloc(int *p)
{
p=(int*)malloc(10*sizeof(*p));
}
main()
{
int *p=NULL;
alloc(p);
if(NULL==p)
printf("alloc failed");
else
printf("alloc successful");
}
Ans: alloc failed
make changes to get “alloc successful” as output.
2. what is the output of the program ?
#include<stdio.h>
main()
{
struct node
{
int a;
int b;
int c;
};
struct node s={3,5,6};
struct node *pt=&s;
printf("%d",*(int*)pt);
}
Ans: 3
3. what is the output of the following program ?
#include<stdio.h>
main()
{
printf(3+"Hello World");
}
Ans: lo World
4. what is the output of the following program ?
#include<stdio.h>
main()
{
int a=0;
if(a=2)
printf("I am inside IF\n");
printf("It is correct");
}
Ans:
I am inside IF
It is correct
5. what is the output ?
#include<stdio.h>
main()
{
unsigned i=5;
if(i > -5)
printf("it is less than -5");
}
Ans: No output
