c++ - For 循环不递增或 while 循环不失败测试

Question

我有这部分代码应该找到给定范围内所有整数的 Collat​​z 序列，该范围由用户输入定义。问题是在 for 循环中，current_number 永远不会递增，或者在内部 while 循环中，current_number != 1 永远不会失败。我错过了什么？

while (lower != 0 && upper != 0) {
    cout << "Lower bound (integer): ";
    cin >> lower;
    cout << "Upper bound (integer): ";
    cin >> upper;
    if (lower == 0 || upper == 0)
        return 0;
    for (current_number = lower; current_number <= upper;
        ++current_number) {
            //cout << current_number << endl;
            counter = 0;
            sequence = sequence + to_string(current_number) + ", ";
            while (current_number != 1) {
                if (current_number % 2 == 0) {
                    current_number = current_number / 2;
                    sequence = sequence + to_string(current_number) + ", ";
                }
                else {
                    current_number = current_number * 3 + 1;
                    sequence = sequence + to_string(current_number) + ", ";
                }
                cout << current_number << endl;
                ++counter;
            }
            //if (counter > longest) {
            //  longest = counter;
            //  the_longest_seed = current_number;
            //}
    }
    cout << sequence << endl;

}

Answer 1

current_number % 2 == 0 对于所有 current_number = 0, 2, 4, 6, ... 都是真的。

发生这种情况时，您将 current_number 设置为 current_number/2。因此，当 current_number 为 2 时，您将其设置为1，然后你递增它 (++current_number)，然后它又是 2 (!= 1)，所以你将输入 while循环，然后因为 2 % 2 = 0 你会再次将它设置为 1.. 依此类推.. :_(

生活小贴士:调试您的代码，这会节省时间、精力，有时还省钱。