查找偶数/奇数，而不使用数学/按位运算符 [英] Find Even/Odd number without using mathematical/bitwise operator

问题描述

我最近被要求编写一个程序，它决定一个数字是偶数还是奇数，而不使用任何数学/位运算符！

任何想法？ >

谢谢！

解决方案

这可以使用1位字段

 ＃include< iostream> 
 
 struct OddEven {
 unsigned a：1; 
}; 
 int main（）{
 int num; 
 std :: cout<<输入数字：; 
 std :: cin>> num; 
 OddEven obj; 
 obj.a = num; 
 if（obj.a == 0）
 cout<<Even！; 
 else 
 cout<<Odd！; 
 return 0; 
} 
  

由于obj.a是单字段值，举行那里！你可以检查你的答案.. 0 - >即使其他奇数.. !!

I was recently asked to write a program, that determines whether a number is even or odd without using any mathematical/bitwise operator!

Any ideas?

Thanks!

解决方案

This can be done using a 1 bit field like in the code below:

#include<iostream>

struct OddEven{
    unsigned a : 1;
};
int main(){
    int num;
    std::cout<<"Enter the number: ";
    std::cin>>num;
    OddEven obj;
    obj.a = num;
    if (obj.a==0)
        cout<<"Even!";
    else
        cout<<"Odd!";
    return 0;
}

Since that obj.a is a one-field value, only LSB will be held there! And you can check that for your answer.. 0 -> Even otherwise Odd..!!

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