


C program written using pointers to find the type of an array entered by the user
Question
Write a C program to find the array type by pointer we need to check whether a given element in the array is even, odd or both The combination.
Solution
The user must enter an array of integers and then the type of the array is displayed.
Example 1 − Input: 5 3 1, output: odd array
Example 2 − Input: 2 4 6 8, output: even number Array
Example 3 - Input: 1 2 3 4 5, Output: Mixed Array
Algorithm
Refer to the algorithm given below to find users Input array type
Step 1: Read the size of the array at runtime.
Step 2: Enter the array elements.
Step 3: Declare pointer variables.
Step 3: Use pointer variables to check whether all elements of the array are odd numbers.
Then, print "Odd".
Step 4: Use pointer variables to check whether all elements of the array are even numbers.
Then, print "Even".
Step 5: Otherwise, print "Mixed".
>
Example
The following is a C program to find the array type entered by the user through a pointer-
Live demonstration
#include<stdio.h> #include<stdlib.h> int*createArray (int); void readArray(int,int *); int findType(int , int *); int main(){ int *a,n,c=0,d=0; printf("Enter the size of array</p><p>"); scanf("%d",&n); printf("Enter the elements of array</p><p>"); createArray(n); readArray(n,a); findType(n,a); return 0; } int *createArray(int n){ int *a; a=(int*)malloc(n*sizeof(int)); return a; } void readArray(int n,int *a){ for(int i=0;i<n;i++){ scanf("%d",a+i); }} int findType(int n, int *a){ int c=0,d=0; for(int i=0;i<n;i++){ if(a[i]%2==0){ c++; } else{ d++; }} if(c==n){ printf("The array type is Even</p><p>"); } if(d==n){ printf("The array type is Odd</p><p>"); } if(c!=n && d!=n){ printf("The array type is Mixed</p><p>"); } return 0; }
Output
When executing the above program, the following output will be produced-
Enter the size of array 4 Enter the elements of array 12 14 16 18 The array type is Even
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