How do I check if an integer is even or odd using

问题:

How do I check if an integer is even or odd using bitwise operators

回答1:

In C (and most C-like languages)

if (number & 1) {
  // It's odd 
}

回答2:

Consider what being "even" and "odd" means in "bit" terms. Since binary integer data is stored with bits indicating multiples of 2, the lowest-order bit will correspond to 2⁰, which is of course 1, while all of the other bits will correspond to multiples of 2 (2¹ = 2, 2² = 4, etc.). Gratuituous ASCII art:

NNNNNNNN
||||||||
|||||||+−− bit 0, value =   1 (20)
||||||+−−− bit 1, value =   2 (21)
|||||+−−−− bit 2, value =   4 (22)
||||+−−−−− bit 3, value =   8 (23)
|||+−−−−−− bit 4, value =  16 (24)
||+−−−−−−− bit 5, value =  32 (25)
|+−−−−−−−− bit 6, value =  64 (26)
+−−−−−−−−− bit 7 (highest order bit), value = 128 (27) for unsigned numbers,
                 value = -128 (-27) for signed numbers (2's complement)

I've only shown 8 bits there, but you get the idea.

So you can tell whether an integer is even or odd by looking only at the lowest-order bit: If it's set, the number is odd. If not, it's even. You don't care about the other bits because they all denote multiples of 2, and so they can't make the value odd.

The way you look at that bit is by using the AND operator of your language. In C and many other languages syntactically derived from B (yes, B), that operator is &. In BASICs, it's usually And. You take your integer, AND it with 1 (which is a number with only the lowest-order bit set), and if the result is not equal to 0, the bit was set.

I'm intentionally not actually giving the code here, not only because I don't know what language you're using, but because you marked the question "homework." :-)

回答3:

if (number & 1)
    number is odd
else // (number & 1) == 0
    number is even

For example, let's take integer 25, which is odd. In binary 25 is 00011001. Notice that the least significant bit b0 is 1.

00011001    
00000001   (00000001 is 1 in binary)
       &
--------
00000001

回答4:

Just a footnote to Jim's answer.

In C#, unlike C, bitwise AND returns the resulting number, so you'd want to write:

if ((number & 1) == 1) {
   // It's odd
}

回答5:

if(x & 1)                               // '&' is a bit-wise AND operator
    printf("%d is ODD\n", x);
else
    printf("%d is EVEN\n", x);

Examples:

    For 9:

      9 ->        1 0 0 1
      1 ->     &  0 0 0 1
      -------------------
      result->    0 0 0 1

So 9 AND 1 gives us 1, as the right most bit of every odd number is 1.

     For 14:

       14 ->      1 1 1 0
       1  ->   &  0 0 0 1
       ------------------
       result->   0 0 0 0

So 14 AND 1 gives us 0, as the right most bit of every even number is 0.

回答6:

You can do it simply using bitwise AND & operator.

if(num & 1)
{
    //I am odd number.
}

Read more over here - Checking even odd using bitwise operator in C

回答7:

Also in Java you will have to use if((number&1)==1){//then odd}, because in Java and C# like languages the int is not casted to boolean. You'll have to use the relational operators to return a boolean value i.e true and false unlike C and C++ like languages which treats non-zero value as true.

回答8:

#include <iostream>
#include <algorithm>
#include <vector>

void BitConvert(int num, std::vector<int> &array){
    while (num > 0){
        array.push_back(num % 2);
        num = num / 2;
    }
}

void CheckEven(int num){
    std::vector<int> array;
    BitConvert(num, array);
    if (array[0] == 0)
        std::cout << "Number is even";
    else
        std::cout << "Number is odd";
}

int main(){
    int num;
    std::cout << "Enter a number:";
    std::cin >> num;

    CheckEven(num);
    std::cout << std::endl;

    return 0;
}