Searching an element in an Array using Recursive Function in C Language - c

Question

I am a beginner to arrays in recursion so need some guidance.
I am trying to find whether an element is present in an array or not.
// Program to find whether an element exist in an array or not.
#include <stdio.h>
int arr[5]= {1,2,3,4,5};
int fooSearch(int array1[],int N,int i, int X)
{
if(i==N)
return 0;
else if (array1[i]==X)
return 1;
else
return fooSearch(array1,N,i++,X);
}
// N denotes total size 5
// i counter that moves from 0 to 4 and eliminate recursion when it reaches 5
// X is the element to be found
int main() {
fooSearch(arr,5,0,3);
return 0;
}
The error I obtained is Segmentation Fault (SIGSEGV).
Please guide me what wrong I am doing with this code.

Answer 1

i++ is a post-fix increment, which increments i after the expression containing it is evaluated. Thus, every call to fooSearch effectively becomes to fooSearch(array1, N, 0, X). Recursion is endless, hence the segfault (or a stack-overflow on my compiler). (You can confirm that i is unchanging by placing printf("%d\n", i) at the top of the function.)
Fix this by using pre-fix increments, which increment the variable before evaluation.
return fooSearch(array1, N, ++i, X);
Or use i+1, since you won't be reusing the local variable anyways.
return fooSearch(array1, N, i+1, X);

Answer 2

While calling fooSearch() recursively pass i+1 instead of i++ as post increment i++ doesn't change i in the argument. For e.g
fooSearch(array1,N,i+1,X);

Answer 3

to find whether an element is present in an array or not.
you can initialize number of array elements, Let's say 10 elements:
int num[10]= {2,3,5,6,1,8,4,9,0,7};
Then, Creating for loop for checking if number 9 is not in array then continue until the condition is false then print the element location.
for(i=0; i<10; i++){
if(num[i] != 9){
continue;
}
printf("9 is found here\n%d",i);
break;
}
At the end, You write an if condition to check if the loop is ended and print not found.
if(i==10){
printf("Not Found");
}
The full code is here:
#include <stdio.h>
int num[10]={2,3,5,6,1,8,4,9,0,7};
int i;
int main(void){
for(i=0; i<10; i++){
if(num[i] != 9){
continue;
}
printf("9 is found here\n%d",i);
break;
}
if(i==10){
printf("Not Found");
}
getchar();
return 0;
}

Answer 4

You have to use ++i or i+1 instead of i++.
i++ is a post increment operator, so the value of i that will go to the function will not change.
Just use ++i or i+1 and you will get the answer.

Answer 5

public static boolean checkNumber(int input[], int x) {
int n=input.length;
if(n==0)
return false;
if(input[0]==x)
{
return true;
}
int small[]=new int[n-1];
for(int i=1;i<n;i++)
{
small[i-1]=input[i];
}
return checkNumber(small, x);
}

Answer 6

// Program to find whether an element exist in an array or not.
// Number of elements of array is N, the number to search is X
#include <stdio.h>
int arr[]= {1,2,3,4,5};
int fooSearch(int array1[],int N,int i, int X)
{ if(i==N)
return 0;
else if (array1[i]==X)
return 1;
else
i=i+1;
return fooSearch(array1,N,i,X);
}
int main() {
int x = fooSearch(arr,5,0,9);
printf("%d",x);
return 0;
}