为什么我的功能不能正常工作

Question

我检查向量中的数字是偶数还是奇数的函数无法正常工作。此函数打印我在向量中键入的数字并将它们放入 2 个类别 EVENS/ODDS。但是，如果我有一个负数，那么不打印它但均匀工作的赔率有问题。有问题的代码在这里：

void printEvensAndOddsVector(const vector <int>& new_v1)
{
    cout << "Vector Evens: ";
    for (unsigned int i = 0; i < new_v1.size(); ++i)
    {
        if (new_v1.at(i) % 2 == 0)
        {
            cout << new_v1.at(i) << " ";
        }
    }
    cout << " \n";

    cout << "Vector Odds: ";
    for (unsigned int i = 0; i < new_v1.size(); ++i)
    {
        if (new_v1.at(i) % 2 == 1) // Here is the problem.
        {
            cout << new_v1.at(i) << " ";
        }
    }
    cout << " \n";
}

或者更准确地说是这一行：if (new_v1.at(i) % 2 == 1) 它打印奇数，但只打印正数而不打印负数。但是，如果我将其更改为 if (new_v1.at(i) % 2 != 0)，则它可以正常工作。为什么会发生这种情况，等于运算符是否存在问题？如果是，那么为什么在仍然使用等于运算符的情况下打印均匀值，即使它们是负数？

此处代码供参考

clude <iostream>
#include <vector>

using namespace std;

void fillVector(vector <int>&);

void printVector(const vector <int>&);

void printEvensAndOddsVector(const vector <int>&); // Prints Evens and Odds.

int main()
{
    vector <int> v1;

    fillVector(v1);
    printVector(v1);
    printEvensAndOddsVector(v1);

    cout << "\n\n\n";
    system( "pause");
    return 0;
}

// Function Definitions
void fillVector(vector <int>& new_v1)
{
    int number;
    cout << "Type in numbers and type -100 to stop: ";
    cin >> number;

    while (number != -100)
    {
        new_v1.push_back(number);
        cin >> number;
    }
}

void printVector(const vector <int>& new_v1)
{
    cout << "\nVector: ";
    for (unsigned int i = 0; i < new_v1.size(); ++i)
    {
        cout << new_v1.at(i) << " ";
    }
    cout << " \n";
}

void printEvensAndOddsVector(const vector <int>& new_v1)
{
    cout << "Vector Evens: ";
    for (unsigned int i = 0; i < new_v1.size(); ++i)
    {
        if (new_v1.at(i) % 2 == 0)
        {
            cout << new_v1.at(i) << " ";
        }
    }
    cout << " \n";

    cout << "Vector Odds: ";
    for (unsigned int i = 0; i < new_v1.size(); ++i)
    {
        if (new_v1.at(i) % 2 == 1)
        {
            cout << new_v1.at(i) << " ";
        }
    }
    cout << " \n";
}

Answer 1

This line: if (new_v1.at(i) % 2 == 1). It prints the numbers that are odd but only the positive ones not the negative ones.

对于 C++ 中的负数 n、n % 2 returns 0 或 -1。它永远不会 returns 1，因此对于负输入，条件不可能成立。

如您所见，与零进行比较是可行的。

见Modulo operator with negative values

为什么我的功能不能正常工作

Why is my function not working properly

c++

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