I want to find a count of odd numbers and even number in a 20 element array using function and this is my attempt

Question

#include<stdio.h>
int oddeven(int i);
int main(){
printf("please enter 20 numbers (without decimal)\n");
int x[20];
int i,search;
for (i = 0; i < 20; i++)
{
printf("Enter the number %d \n",i);
scanf("%d", &x[i]);
}
printf("Enter a number to search\n");
scanf("%d", &search);
for(i=0;i<20;i++){
if(x[i]==search)
printf("%d location %d\n",search,i+1);
}
int k=printf("%d",x[i]);

}
int oddeven(int i)

{ int size,odd_count,even_count,x[i];
for(i = 0; i < size; i++)

if(x[i] % 2 == 1)
odd_count++;
else
even_count++;
}

Answer 1

Oh, now I get it.
You can add a flag variable. Think of it like a boolean that is 0 if the search is unsuccessful and 1 if it succeeded. So, after the loop you check this variable. For example:

int found = 0;
...
for(i=0;i<20;i++){
if(x[i] == search) {
printf("found it") ;
found = 1;
}
if(found==0)
printf("Sorry") ;

Answer 2

There are a few issues with the function. Firstly, the array x is local (in the main function), so the function oddeven won't be able to see it (out of scope). To be able to return both the odd count and the even count, you need to have as input some pointers to integers and change their value in the function.
An implementation would look something like this:

void oddeven(int* array, int size, int* odd_count, int* even_count) {
int i;
for(i=0; i<size; i++) {
if(array[i] % 2 == 1)
*odd_count++;
else
*even_count++;
}
}

To call this function, you would write (in main) :
int odd, even;
oddeven(x, 20, &odd, &even);
The results would be stored in variables odd and even

Answer 3

Thank you and I recode and show you

Answer 4

The problem is the if-else statement. If you have a positive odd number, the odd counter won't be updated, because the c1 counter will be increased and the other cases (else) won't be tested. You can change it like this:

if(x[i] >0)
c1++;
else if(x[i] <0)
c2++;
else
c3++;

if(x[i] % 2 == 0)
c4++;
else
c5++;

Answer 5

Thank you very much it works properly now

Answer 6

You are welcome. Good luck on your coding journey

Answer 7

You can add something like: printf("Sorry\n") ; outside of the loop. What do you mean by count searching result?

Answer 8

I have a search unit in my code it shows the location of the particular no if i want to count the locations of particular no what i van do ?

Answer 9

I recoded it and but i unable to print that oddvount amd even count if i want to print the count shoud i use printf to &odd or normal odd?

Answer 10

#include<stdio.h>
void count(int x[]);
int main(){
printf("please enter 20 numbers (without decimal)\n");
int x[20];
int i,search;
for (i = 0; i < 20; i++)
{
printf("Enter the number %d \n",i);
scanf("%d", &x[i]);
}
printf("Enter a number to search\n");
scanf("%d", &search);
for(i=0;i<20;i++){
if(x[i]==search)
printf("%d location %d\n",search,i+1);
if(x[i]!=search)
printf("sorry\n");
}

count(x);

}
void count(int x[])
{
int i,c1=0,c2=0,c3=0,c4=0,c5=0;
for(i=0;i<20;i++)
{
if
(x[i]>0)
c1++;
else if
(x[i]%2==0)
c4++;
else if(x[i]%2==1)
c5++;
else
if(x[i]<0)
c2++;
else
c3++;

}

Answer 11

In above code i want to print sorry one time and it not works in odd numbers

Answer 12

And one last thing cannt i count searching result or print sorry once ?

Answer 13

Okkkkkkk thank you 😍😍😍😍😍