c++ - 计算 rng 总数时出现意外结果

Question

我编写了一个简单的程序来模拟掷 4 面骰子并计算每个数字出现的次数，这样我就可以看到 rand() 的随机性。当我简单地打印结果时，出现的唯一数字是预期的数字 1->4。当我进行计数时，我注意到在 1000 次迭代中，我最终只得到 1 到 4 之间的约 700 个刻度。为什么会这样？我觉得答案应该很明显，因为循环按预期运行，但是当我查看代码的可疑问题区域时，我看不出问题出在哪里。

int d4()
{
    int x;
    x = rand() % 4 + 1;
    return x;
}  


int main()
{
    int n1 = 0;
    int n2 = 0;
    int n3 = 0;
    int n4 = 0;
    int n5 = 0;
    srand(time(0));
    int x = 0;
    while (x <= 100)
    {

        std::cout << d4() << " ";
        ++x;

/*      ++x;
        if (d4() == 1)
            ++n1;
        else if (d4() == 2)
            ++n2;
        else if (d4() == 3)
            ++n3;
        else if (d4() == 4)
            ++n4;
        else
            ++n5;           */
    }

    std::cout << "Number 1 was rolled " << n1 << " times." << std::endl;
    std::cout << "Number 2 was rolled " << n2 << " times." << std::endl;
    std::cout << "Number 3 was rolled " << n3 << " times." << std::endl;
    std::cout << "Number 4 was rolled " << n4 << " times." << std::endl;
    std::cout << "Somehow some other number was rolled on a 4 sided die " << n5 << " times.";

    std::cin.ignore();
    std::cin.clear();
    std::cin.get();

    return 0;
}

注意:我已经注释掉了计数片段以手动查看结果。

Answer 1

正如其他人在评论中指出的那样，您的问题是每次调用 d4() 时，它都会在 1 和 4< 之间返回不同的结果(包括)。这可能意味着在您的循环中，d4() 的返回值永远不会等于比较时的比较值。考虑:

    if (d4() == 1) // call d4(), which returns 2
        ++n1;
    else if (d4() == 2) // call d4() again, which returns 3
        ++n2;
    else if (d4() == 3) // call d4() again, which returns 1
        ++n3;
    else if (d4() == 4) // call d4() again, which returns 2
        ++n4;
    else
        ++n5; // No other comparison worked, so this is incremented instead.

这里的解决方案是只调用一次d4()并将它的值保存在一个变量中。

// Retrieve the random number this loop iteration and store its value
int randomNumber = d4();
std::cout << randomNumber << std::endl;

if (randomNumber == 1)
    ++n1;
else if (randomNumber  == 2)
    ++n2;
else if (randomNumber  == 3)
    ++n3;
else if (randomNumber  == 4)
    ++n4;
else
    ++n5;