Problem in Loop!!

[fredsilvester93]

n numbers are entered from the keyboard into an array. Write a program to find out many of them are positive, how many are negative, how many are even and how many odd.

Sample Input #Supposed to be)
Enter 10 numbers : 10 20 23 -6 52 34 22 66 -22 -1
Sample Output #1:
Positive Numbers : 7
Negative Numbers : 3
Odd Numbers : 2
Even Numbers : 8

#include <stdio.h>
#define MAX_INP 10

int main ()
{
    int no[MAX_INP]={0},i=0,j=1,pos=0,neg=0,odd=0,even=0;
    
    for (i =0; i<10; i++){
        printf("Enter no %d: ",j);
        scanf("%d",&no[i]);
        j++;
        }
        
    
     for (i=0; i<10; i++){
        if(no[i] > 0){
                 pos++;
                 }
                 
        if(no[i] < 0){
             neg++;
             }
                     
        if((no[i]/2) == 1 || no[i]/2 == -1){
             odd++;
             }
             
        if((no[i]/2) == 0){
             even++;
             }
                         }
                         
        printf("\n\nPositive numbers : %d\nNegative Numbers : %d\nOdd Numbers : %d\nEven Numbers : %d",pos,neg,odd,even);
    
    
    
    getch ();
    return 0;
}

Out put is not shown correctly in case of odd and even while =ve and -ve numbers are showing properly. Please Help!!!!!

[grumpy]

Nothing to do with the loop. It is your tests for what is odd or even that are broken.

If no[i] is 23, then no[i]/2 is 10 (integer arithmetic). The tests you are doing will not classify it as odd or even.

If no[i] is -1 then no[i]/2 is zero. Your test means it will be counter it as even.

[fredsilvester93]

OOOH!!! my mistake. i mistakenly used / instead of %

[CommonTater]

Out put is not shown correctly in case of odd and even while =ve and -ve numbers are showing properly. Please Help!!!!!

The easiy way to detect an odd or even number is by testing bit 0...

if (number & 1)
  odd++;
else
  even++;

[grumpy]

The easiy way to detect an odd or even number is by testing bit 0...

if (number & 1)
  odd++;
else
  even++;

The slightly less cryptic way is use modulo 2. It requires no more typing, and there is some advantage to understanding through computing a (trivial) numeric attribute using a numerical operation rather than a bitwise operation. No need to know what a bit is, what bit 0 is .....

Bitwise operations do have some implementation-defined or undefined characteristics on signed types too.

[fredsilvester93]

The easiy way to detect an odd or even number is by testing bit 0...

if (number & 1)
  odd++;
else
  even++;

Didn't get the Logic... what does ''if (number & 1) means?''


i used ''if(no%2 == 1)''

help plz///

[laserlight]

what does ''if (number & 1) means?

You could read up on bitwise operators, but for your purposes what you used works just as well.

[CommonTater]

Didn't get the Logic... what does ''if (number & 1) means?''


i used ''if(no%2 == 1)''

help plz///

It's capitalizing on how integers are stored in C ... an integer is just a binary number.

In the binary numbering system all odd numbers have a 1 at bit position 0.

0 = 00000000
1 = 00000001 <-- odd
2 = 00000010 < -- even
3 = 00000011 <--- odd
and so on...

So if we AND the number with 1 (i.e. bit 0 = 1) we will get a 1 or a 0 indicating odd or even.

01010011
AND
00000001
======
00000001 <-- odd

11011000
AND
00000001
======
00000000 <-- even

Since an if() statement responds to TRUE (1) or FALSE (0) we can easily use the result of number AND 1 to trigger it.

[fredsilvester93]

It's capitalizing on how integers are stored in C ... an integer is just a binary number.

In the binary numbering system all odd numbers have a 1 at bit position 0.

0 = 00000000
1 = 00000001 <-- odd
2 = 00000010 < -- even
3 = 00000011 <--- odd
and so on...

So if we AND the number with 1 (i.e. bit 0 = 1) we will get a 1 or a 0 indicating odd or even.

01010011
AND
00000001
======
00000001 <-- odd

11011000
AND
00000001
======
00000000 <-- even

Since an if() statement responds to TRUE (1) or FALSE (0) we can easily use the result of number AND 1 to trigger it.

ok . like if (no & 1 == 0) will do even++ of no is 2????

[CommonTater]

ok . like if (no & 1 == 0) will do even++ of no is 2????

Any even number. That's the whole point... that last bit is 1 for all odd numbers, 0 for all even numbers.
The thing is we have to strip the rest off so we can look at that one bit in isolation, so we use AND.

[fredsilvester93]

oooooh gr8... got the concept after applying.. thanks alot allllllll........ it a gr8 concept....