Interview Questions

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Interview Questions

Sasken Technology - Interview Experience

SASKEN Technology- Interview Questions and Answer


Find Output of the Following C Program


Program 1

	#define cal(x) (x*x*++x)
	main()
	{
		int i,j=5;
		i=cal(j);
		printf("%d..%d\n",i,j);
	}


Program 2

	main()
	{
		int i=400,j=800;
		printf("%d..%d");
	}


Program 3

	main()
	{
		int x=10;
		int y=+x++x;
		printf("%d\n",y);
	}


Program 4

	void fun(int n)
	{
		if(n>0)
		{
			fun(--n);
			printf("%d",n);
			fun(--n);
		}
	}
	main()
	{
		int a=3;
		fun(a);
		printf("\n");
	}


Program 5

	void test(void)
	{
		int j=10;
		printf("%d ",j);
		j++;
	}
	main()
	{
		int i;
		for(i=0;i<5;i++)
			test();
	}


Program 6

	main()
	{
		int num;
		num=321;
		num=fu(num);
		printf("%d",num);
	}
	int fu(int num)
	{
		static int n=0;
		int r;
		if(num)
		{
			r=num%10;
			n=n*10+r;
			fu(num/10);
			return n;
		}
		else
			return n;
	}


Program 7

	void holy(int c);
	main()
	{
	  int index;
	  index=3;
	  holy(index);
	}
	void holy(int c)
	{
	  c--;
	  printf("%d",c);  
	  if(c>0)
	    holy(c);
	  printf("%d",c);
	}


Program 8

	#define sq(x) x*x
	main()
	{
		int i;
		i=64/sq(4);
		printf("%d\n",i);
	}
	

Program 9

	char* f(int *i,float *j)
	{
		 static char *s="function  pointer";
		s=s+*i+(int)*j;
		return s;
	}
	main()
	{
		char  *str;
		int x=2;
		float y=3.01;
		char *(*p) (int*,float*);
		p=&f;
		str=(*p)(&x,&y);
		printf("%s",str);
	}


Program 10

	main()
	{
		int a=10,*j;
		void *k;
		j=k=&a;
		j++;
		k++;
		printf("%u %u\n",j,k);
	}


Program 11

	main()
	{
		int a[5]={1,2,3,4,5};
		printf("%u\n",a);// 100 
		printf("%u %u \n",a+1,&a+1);// 104  120
	}


Program 12

	main()
	{
		short a[4][3]={ {1},{2,3},{4,5,6}};
		printf("%d\n",sizeof(a));
	}


Program 13

	main()
	{
		int a[]={1,2,3,4};
		int b[a[3]]={1,2};
		for(int i=0;i<4;i++)
			printf("%d\n",++b[i]);
	}


Program 14

	main()
	{
		int a=5;
		int *b;
		b=&a;
		printf("%d\n",b);
	}


Program 15

	main()
	{
		int *num;
		num=f(45,39);
		printf("%d\n",num);
	}
	int *f(static int i,static int j)
	{
	}


Program 15
	
	main()
	{
		char *str="alice";
		int i=0;
		while(*str !='\0')
		{
			i++;
			str++;
		}
		printf("%d\n",i);
	}


Program 17

	main()
	{
		char str[]="take care";
		f(str);
	}
	void f(char *str)
	{
		printf("%s\n",str);
	}


Program 18

	main()
	{
		char **ptr;
		char *str="hello";
		int i=0;
		ptr=(char**)calloc(sizeof(char**),3);
		if(ptr==0)
			exit(-1);
		while(ptr!=0)
			*(str+i++)=(str+i++);
		for(i=0;i<3;i++)
			printf("%s\n",ptr[i]);
	}


Program 19

	main()
	{
		char emp[45];
		gets(emp);
		puts(emp);
	}


Program 20

	main()
	{
		printf(5+"life is beautiful");
	}


Program 21

	char *f()
	{
		 char *temp="local string";
		return temp;
	}
	main()
	{
		puts(f());
	}
	

Program 22

	#include<stdio.h>
	#define buf 100
	main()
	{
		char b[buf];
		while(fgets(b,buf,stdin)!=NULL)
			printf("%s\n",buf);
	}


Program 23

	main()
	{
		fprintf("action");
		printf("%.ef",2.0);
	}
	
Program 24

A Single Linked list Contains the binary values in each node. Print the corresponding Decimal value for the same

	
	#include<malloc.h>
	#include<stdio.h>
	struct st
	{
		unsigned long long int num;
		struct st *next;
	};
	struct st *hptr;
	void add();
	void print();
	main()
	{
		int i,node;
		printf("enter the number of nodes you want to print\n");
		scanf("%d",&node);
		for(i=0;i<node;i++)
			add();
		print();
	}
	void add()
	{
		struct st *temp;
		temp=malloc(sizeof(struct st ));
		printf("enter the binary number\n");
		scanf("%llu",&temp->num);
	
		temp->next=hptr;
		hptr=temp;
	}
	void print()
	{
		struct st *temp=hptr;
		int n,rem,j=1,sum=0;
		while(temp)
		{
				n=temp->num;
				while(n)
				{
					rem=n%10;
					sum=sum+rem*j;
					j=j*2;
					n=n/10;
				}
				printf("[%llu] equivalent decimal is : [%d]\n",temp->num,sum);
				temp=temp->next;
				sum=0;
				j=1;
			}
	}
	
	
Program 25
	
  Find the lowest path from the following
	
      1  5  8
      6  7  9
     12 10  11
	
  Following condition
  
     a)  Start From A(0,0)
     b)  Reach till A(2,2)
     c)  Traverse in such a way that each time Take One Left and then Take 
         One Right or Take one Right then take one left and reach till A(2,2)
     d)  Sum the values at each index traversed and find the lowest path.
	
	#include<stdio.h>
	main()
	{
		int a[3][3]={1,5,8,6,7,9,12,10,11};
		int i,j,c=0,k,sum=0;
		for(i=0,j=0;i<3;c++)
		{
	
			sum=sum+a[i][j];
			printf("%d ",a[i][j]);
			if(c%2==0)
			{
				printf(" ");
				i++;
			}
			else
			{
				printf("\t");
				j++;
			}
	
		}
		printf("sum=%d\n",sum);
	}


Program 26
	
   /*  There are 3 classes, Alpha, Beta, Gama
	
	   a.  Alpha and Beta are predefined classes
	   b.  Gama derived from Alpha, Beta
	   c.  u and v are class members of Alpha,
	   d.  Create constructor in Gama class with parameters a and b
	   e.  Assign values of a and b to u and v
	   f.  show() function will be called from Gama Class and values will 
	        be displayed.
    */
	
	
	#include<iostream>
	using namespace std;
	class alpha
	{
		private:
			int u;
			int v;
		public:
			int show()
			{
				cout<<"in alpha_show()...."<<endl;
				cout<<"u= "<<u<<endl;
				cout<<"v= "<<v<<endl;
			}
			alpha(int x,int y)
			{
				u=x;
				v=y;
			}	
	
	};
	class beta
	{
		private :
	
	};
	class gama : public alpha,public beta
	{
	
		public :
		gama(int a,int b):alpha(a,b)
		{
		}
	};
	main()
	{
		gama g(10,20);
		g.show();
	}


Program 27
	
/* . Jump N Jacks :

	Jack starts at 0 on the number line and jumps one unit in either direction. Each successive jump he makes is longer than the previous by 1 unit, and can be made in either direction. Write a program that takes a number and returns the minimum number of jumps Jack makes to reach that number.
	
For example,
	It takes seven jumps to reach 12, first a jump left to -1, then a jump right to 1, a jump right to 4, a jump right to 8, a jump right to 13, a jump right to 19, and a jump left to 12. To go to -12, simply reverse the direction of each jump.
	Your task is to write programs that compute the minimum number of steps Jack must take to reach his target and that compute the actual steps.   */
	
	
	#include <stdio.h>
	#include <limits.h>
	#include<malloc.h>
	int min(int x, int y) 
	{ 
		return (x < y)? x: y; 
	}
	
	// Returns minimum number of jumps to reach a[n-1] from a[0]
	int min_jumps(int a[], int n)
	{
		int *p =malloc(100);  //p[n-1] will hold the result
		int i,j;
	
		if(n==0 || a[0]==0)
			return INT_MAX;
	
		p[0]=0;
	
		// Find the minimum number of jumps to reach a[i]
		// from a[0],and assign this value to p[i]
		for(i=1;i<n;i++)
		{
			p[i]=INT_MAX;
			for(j=0;j<i;j++)
			{
				if(i<=j+a[j]&&p[j]!=INT_MAX)
				{
					p[i]=min(p[i],p[j] + 1);
					break;
				}
			}
		}
		return p[n-1];
	}
	main()
	{
	    int a[]={1, 3, 6, 1, 0, 9};
	    int n=sizeof(a)/sizeof(a[0]);
	    printf("Minimum number of jumps to reach end is %d \n", min_jumps(a,n));
	}
Interview Questions

sasken-technology - Interview Experience

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Interview Questions

Sasken Technology - Interview Experience

SASKEN Technology- Interview Questions and Answer

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