I can't store integers inside an array - c

Question

This is an activity given by my instructor.
Create a program that accepts numeric input from the user. If the user enters an even number, store it to an array for even numbers. If the user enters an odd number, store it to another array for odd numbers. Input terminates if the user entered 10 numbers already. Display the size of each array and their elements.
Example:
Input: 5, 6, 12, 10, 0, 3, 4, 100, -1, 7
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
and this is the code I've come up with.
#include <stdio.h>
int sort(int);
int main(){
int input, count;
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
sort(input);
}
printf("%d", input);
return 0;
}
int sort(int inp){
int odd[10];
int even[10];
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Please help me on how to store the numbers into two separate arrays. Any tips will be greatly appreciated.

Answer 1

Check the below code. It's self-explaining.
#include <stdio.h>
#include <stdlib.h>
#define NUM 10
int main()
{
int input, i;
int oddcounter = 0, evencounter =0;
int oddarr[NUM];
int evenarr[NUM];
printf("Enter 10 integers\n");
for (i = 0; i < NUM; i++)
{
if ( scanf("%d", &input) == 1 )
{
if ((input % 2) == 0)
{
evenarr[evencounter++] = input;
}
else
{
oddarr[oddcounter++] = input;
}
}
}
printf("Number of elem in oddarray : %d, evenarray : %d\n\n", oddcounter, evencounter);
printf("Odd elements are :");
for (i = 0; i < oddcounter ; i++) printf("%d\t", oddarr[i]);
printf("\n");
printf("Even elements are :");
for (i = 0; i < evencounter; i++) printf("%d\t", evenarr[i]);
printf("\n");
return 0;
}

Answer 2

In addition to Sourav's comment which indicates that you shouldn't have the int[] arrays be local to sort, this syntax isn't correct for assigning to arrays in C:
odd[]=inp;
On my compiler, it generates the following error:
24:9: error: expected expression
odd[]=inp;
To store to odd, you need to indicate the index at which you'd like to store, for example:
odd[1]=inp;
which also means you'll need to keep track the latest index you wrote to for each array!

Answer 3

You need to tell the compiler which index of the array you are storing your data to. In sort:
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Should look something like:
if(inp%2==0){
odd[endofoddindex]=inp;
}
else
even[endofevenindex]=inp;
return 0;
}
That said, you won't get much use out of the arrays being local variables, since they are deallocated on each call. Your best bet is to declare the arrays in main and pass them in.

Answer 4

Your even and odd arrays are both local. This means that they exist as long as the function exists. So you won't be retrieve the data you have stored(You also don't store it correctly).
So you need both the arrays in main and also two other variables for using as the index of both the array(i and j in the below program). The modified program is given below:
#include <stdio.h>
int sort(int);
int main(){
int input, count,i=0,j=0; //i and j to be used as array indices
int odd[10];
int even[10]; //arrays in main
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
if(sort(input)) //if odd number was found
odd[i++]=input;
else //even number found
even[j++]=input;
}
printf("%d", input);
//print even and odd arrays here
return 0;
}
int sort(int inp){
if(inp%2==0)
return 0; //if number is even,return 0
return 1; //else return 1
}

Answer 5

You need to either have your arrays as globals or pass them into your sort function. Where they are they currently they get recreated every time the sort function is called and are inaccessible to the rest of your program.
You will also need to keep track of the max number of ints in each array and the current number.
Your test in sort would be something like this:
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ currentEvenCount ] = inp;
currentEvenCount++
}
else
{
//TODO check that currentOddCount < maxOddCount
odd[ currentOddCount ] = inp;
currentOddCount++;
}
To declare your arrays as globals just move the declaration outside of any function above anywhere they are referenced
int even[10];
int odd[10];
int main() ...
To pass them as parameters to sort function you could declare sort like this:
sort( int inp, int even[], int maxEvenCount, int* currentEvenCount, int odd[]. int maxOddCount, int* currentOddCount)
{
...
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ *currentEvenCount ] = inp;
(*currentEvenCount)++
}
}
The * in front of currentEventCount is dereferencing the pointer and getting/setting the actual value pointed to.
You would then call sort like so:
int main()
{
int evenArray[10];
int oddArray[10];
int currentEvenCount = 0;
int currentOddCount = 0;
...
sort( input, evenArray, 10, &currentEvenCount, oddArray, 10, &currentOddCount);
}

Answer 6

There is no any sense to define the arrays as local variables of function sort because each time the function is called the arrays are created anew.
The program could look the following way
#include <stdio.h>
#define N 10
enum Type { Even, Odd };
enum Type sort( int x )
{
return x % 2 == 0 ? Even : Odd;
}
int main( void )
{
int odd[N];
int even[N];
int odd_count = 0;
int even_count = 0;
int i;
printf( "Enter %d numbers: ", N );
for( i = 0; i < N; i++ )
{
int num;
scanf( "%d", &num );
switch ( sort( num ) )
{
case Even:
even[even_count++] = num;
break;
case Odd:
odd[odd_count++] = num;
break;
}
}
printf( "Even numbers (%d):", even_count );
for ( i = 0; i < even_count; i++ ) printf( " %d", even[i] );
printf( "\n" );
printf( "Odd numbers (%d):", odd_count );
for ( i = 0; i < odd_count; i++ ) printf( " %d", odd[i] );
printf( "\n" );
return 0;
}
If to enter
5 6 12 10 0 3 4 100 -1 7
then the output will be
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
Simply copy, paste and investigate the program.:)

Answer 7

Hope this program will solve you issue. Here is the working code.
#include <stdio.h>
int sort(int[]);
int main(){
int input[10], count;
printf("Enter 10 digits: ");
for(count=0;count<10;count++){
scanf("%d", &input[count]);
}
sort(input);
return 0;
}
int sort(int inp[]){
int odd[10];
int even[10];
int oddCount=0, evenCount=0;
int i;
for(i=0; i<10;i++)
{
if(inp[i]%2==0){
even[evenCount]=inp[i];
evenCount++;
}
else
{
odd[oddCount]=inp[i];
oddCount++;
}
}
printf("ODD COUNT is %d \n", oddCount);
for(i=0; i<oddCount;i++)
{
printf("ODD VALUE %d \n", odd[i]);
}
printf("EVEN COUNT is %d \n", evenCount);
for(i=0; i<evenCount;i++)
{
printf("EVEN VALUE %d \n", even[i]);
}
return 0;
}

Answer 8

Your variable input is just an int, it's not an array. You need two arrays:
int even[10], odd[10];
Then you need to keep track of the number of numbers you've read so far, and for each number check which array to store it in.
I don't see a need to do sorting, so not sure why you have a function called sort().
It should just be something like:
int even[10], odd[10];
int oddindex = 0, evenindex = 0;
while(scanf(" %d", &x) == 1)
{
if(x % 2 == 0)
even[evenindex++] = x;
else
odd[oddindex++] = x;
if((evenindex + oddindex) >= 10)
break;
}
/* Loop here to print numbers. */

Answer 9

An answer suitable for an assignment question:
int main()
{
int i,c,o[10],e[10];int oc=0;int ec=0;int*pc;for(c=0;c<10;c++){scanf("%d",&i);pc=(i&1)?&o[oc++]:&e[ec++];*pc=i;}
// Now print out the values as requested in oc, o, ec and e.
}