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.
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
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".
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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; }
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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