Why is the program print 0 for the number of even elements in the array , I have used faster approac-CodePudding

// here I have tried to print the number of positive , negative ,even , odd , elements in the array . The rest of the output is right , except it is print 0 in case of even , when obviously they are not 0.So What is the bug here!

#include <stdio.h>

int main()
{ 

    int arr[10] = {2, 4, 6, 8, 10, 11, 13, 15, -2, -4};
    int pos = 0;
    int neg = 0;
    int even = 0;
    int odd = 0;

    for (int i = 0; i < 10; i  )
    {
        if (arr[i] & 1 == 0)
        {
            even  ;
        }
        if (arr[i] & 1 == 1)
        {
            odd  ;
        }
        if (arr[i] > 0)
        {
            pos  ;
        }
        else if (arr[i] < 0)
        {
            neg  ;
        }
    }
    printf("the odd are %d ,the even are %d ,the positive are %d , the negative are %d\n", odd, even, pos, neg);

    return 0;
}

CodePudding user response：

This is because of operator precedence

arr[i] & 1 == 0 is equivalent to arr[i] & (1 == 0), which is arr[i] & 0, which is obviously 0, so the condition is always false.

You want to do this instead:

if ((arr[i] & 1) == 0)
{
    even  ;
}
if ((arr[i] & 1) == 1)
{
    odd  ;
}

The odd test already works because arr[i] & 1 == 1 is arr[i] & (1 == 1) which is arr[i] & 1 which is equivalent to what you were checking anyway.

CodePudding user response：

arr[i] & 1 == 0 is always zero because of operator precendence. It is equivalent with arr[i] & (1 == 0) which is clearly not what you want.

Try

if ((arr[i] & 1) == 0) {
    even  
}

CodePudding user response：

& has lower precedence than ==, so arr[i] & 1 == 0 is structure as arr[i] & (1 == 0) instead of (arr[i] & 1) == 0. To get the latter, you must use explicit parentheses.

Turn on warnings in your compiler, elevate them to errors, and pay attention to the compiler messages. With Clang, start with -Wmost -Werror. With GCC, start with -Wall -Werror. With MSVC, start with /W3 /Wx.

CodePudding user response：

The problem with your code is the conditions you're using. They're wrong. The correct way to find out whether a number is odd or even is this:

if (num % 2 == 0)
{
    // Even
}
if (num % 2 != 0)
{
    // Odd
}

So the final code:

#include <stdio.h>

int main()
{

    int arr[10] = { 2, 4, 6, 8, 10, 11, 13, 15, -2, -4 };
    int pos = 0;
    int neg = 0;
    int even = 0;
    int odd = 0;

    for (int i = 0; i < 10; i  )
    {
        if (arr[i] % 2 == 0)
        {
            even  ;
        }
        if (arr[i] % 2 != 0)
        {
            odd  ;
        }
        if (arr[i] > 0)
        {
            pos  ;
        }
        else if (arr[i] < 0)
        {
            neg  ;
        }
    }
    printf("the odd are %d ,the even are %d ,the positive are %d , the negative are %d\n", odd, even, pos, neg);

    return 0;
}