After scanning all array elements how can i printf even and odd seperately.

#include <stdio.h>  
void  main()  
{  
    int arr[100]; 
    int n,i;
    printf("Number of Elements:-");
    scanf("%d",&n);
    for (i=0;i<=n;i=i 1)
    {
        printf("Elem %d:-",i);
        scanf("%d",&arr[i]);
    }
        for(i=0;i<=n;i=i 1)
        {
            if(arr[i]/2==0)
            {
                printf("Even%d\n",arr[i]);
            }
            else
            {
                printf("odd%d\n",arr[i]);
            }
        }

}

Output:

Number of Elements:-4
Elem 0:-1
Elem 1:-2
Elem 2:-3
Elem 3:-4
Elem 4:-5
Even1
odd2
odd3
odd4
odd5

CodePudding user response：

The conventional way to write your loop is:

for( i=0; i<n;   i )
{
    printf("%s %d\n", (arr[i]%2)? "Odd" : "Even", arr[i]);
}

In C, most loops have an initial condition of var=0, run while var<target, and increment with var (some people prefer var )

The conventional way to test if a number is even or odd is using the Modulo (%) operator, not using division (/).

Why do you think a an expression like 10/2 would ever equal 0??
10 is clearly even, dividing by 2 gives 5, not zero!!?

Putting it all together, I got:

#include <stdio.h> 
#include <stdlib.h>

int main(void)  
{  
    int arr[100]; 
    int n = 20;
    
    for ( int i=0; i<n;   i )
    {
        arr[i] = rand() % 100;
        // Replaced user-input with pseudo-random numbers, for ease and simplicity. 
        // You can stick with scanf if you want.
    }
    
    for( int i=0; i<n;   i )
    {
        printf("%-4s %d\n", (arr[i]%2)? "Odd" : "Even", arr[i]);
    }
    
    return 0;
}

Output:

Success #stdin #stdout 0.01s 5404KB
Odd  83
Even 86
Odd  77
Odd  15
Odd  93
Odd  35
Even 86
Even 92
Odd  49
Odd  21
Even 62
Odd  27
Even 90
Odd  59
Odd  63
Even 26
Even 40
Even 26
Even 72
Even 36

CodePudding user response：

main() returns an int unless you are using a Microsoft extension. Use a constant LEN instead of hard-coded magic values. Added boundary checks on your array size, and checked the return value of scanf() to ensure you read a value. Localized the loop variable i and used the common idiom for loops in c for(int i = 0; i < n; i ). The modular operator %, particular, % 2 returns 0 for even and 1 for odd. "separately" implies two loops to me.

#include <stdio.h>
#define LEN 100

int main() {
    int arr[LEN];
    printf("Number of Elements: ");
    int n;
    if(scanf("%d", &n) != 1) {
        printf("scanf failed\n");
        return 1;
    }
    if(n < 1 || n > LEN) {
        printf("n must be between 1 and 100\n");
        return 1;
    }
    for (int i=0; i<n; i  ) {
        printf("Elem %d: ", i);
        if(scanf("%d", &arr[i]) != 1) {
            printf("scanf failed\n");
            return 1;
        }
    }
    for(int i=0; i<n; i  )
        if(arr[i] % 2) printf("odd: %d\n",arr[i]);

    for(int i=0; i<n; i  )
        if(!(arr[i] % 2)) printf("even: %d\n",arr[i]);
}

and here is the corresponding output:

Number of Elements: 4
Elem 0: 0
Elem 1: 1
Elem 2: 2
Elem 3: 3
odd: 1
odd: 3
even: 0
even: 2