Answer / dartz

#include<stdio.h>
#include<conio.h>
void main()
{
int ch[100],m,n,flag=0;
printf("enter the limit value");
scanf("%d",&m);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
printf("enter the number u are searching for :");
scanf("%d",&n);
for(i=0;i<m;i++)
{
if(a[i]==n)
{
printf("%d number is found at the position %d in the
given array ",n,i);
i=m;// to come out from the loop when we found
searching number
flag=1;

}
}


if( flag = 0)
{
printf(" given number is not found in the list of array");

}




getch();
}

Answer / ashwin kumar

dear no need of pointer for the above Question dear

#include<stdio.h>
#include<conio.h>
void main()
{
int ch[100],m,n,flag=0;
printf("enter the limit value");
scanf("%d",&m);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
printf("enter the number u are searching for :");
scanf("%d",&n);
for(i=0;i<m;i++)
{
if(a[i]==n)
{
printf("%d number is found at the position %d in the
given array ",n,i);
i=m;// to come out from the loop when we found
searching number
flag=1;

}
}


if( flag = 0)
{
printf(" given number is not found in the list of array");

}




getch();
}

Answer / vignesh1988i

small mistake in above program.... in last printf pl. change

*(*(ptr+i)) as (*(ptr+i))

Answer / vignesh1988i

here i have used a concpt of pointers .. ie. array of
pointers.... when you search for a number it may occur once
or more than it... i have designed for all cases.. that's
why i used pointers......................

#include<stdio.h>
#include<conio.h>
void main()
{
int ch[100],m,n,*ptr[20],count=0;
printf("enter the limit value");
scanf("%d",&m);
for(int i=0;i<m;i++)
scanf("%d",&a[i]);
printf("enter the number u are searching for :");
scanf("%d",&n);
for(i=0,k=0;i<m;i++)
{
if(a[i]==m)
{
count++;
ptr[k]=&a[i];
k++;
}
for(i=0;i<count;i++)
printf("the number occured %d times and found to be int
these addresses : %u ",count,*(*(ptr+i)));
getch();
}

#include<stdio.h> main() { int a[3]; int *I; a[0]=100;a[1]=200;a[2]=300; I=a; Printf(“%d\n”, ++*I); Printf(“%d\n”, *++I); Printf(“%d\n”, (*I)--); Printf(“%d\n”, *I); } what is the o/p a. 101,200,200,199 b. 200,201,201,100 c. 101,200,199,199 d. 200,300,200,100

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