write a function that accepts an integer/char array and an
search item.If the search item is there in the array return
position of array and value else return -1.without using
other array,without sorting,not to use more than one loop?
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Answer / nikhil srivastav (mca pesit b
int search(int *arr,int item,int arr_size)
{
int i;
for(i=0;i<arr_size;i++)
{
if(item==*(arr+i))
return i;
}
return -1;
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / p govind rao
#include<stdlib.h>
#include<stdio.h>
#define Max 6
int fun_rev(int *ptr, int num)
{
int i=0;
while(i<=Max)
{
if(num==*ptr)
{
return 1 ;
}
else
{
return 0;
}
i++;
ptr++;
}
}
int main()
{
int arr[Max]={3,4,5,6,2,1};
int item=6,result;
result=fun_rev(arr,item);
printf("result = %d",result);
return 0;
}
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / ashutosh tiwari
int find_num(int *arr, int *arr_num, int arr_size)
{
int i;
while((i<arr_size) && (*(arr+i) != *arr_num))
i++;
if(i >= arr_size)
return -1;
else
return i;
}
OR
int find_num(int *arr, int *arr_num, int arr_size)
{
int i;
for(i=0;i<arr_size;i++)
{
if(*(arr+i) != *arr_num)
continue;
else
return i;
}
return -1;
}
input to function will be actual array, number to be found
with its reference and array size
output will be -1 if fail otherwise number position
| Is This Answer Correct ? | 2 Yes | 1 No |
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