迭代后如何从主函数重复执行

Question

所以我有一个简化布尔表达式的程序。我想要实现的是，在第一个表达式的简化结束时，我希望用户选择是要简化另一个表达式还是退出控制台应用程序（程序）。

这是 Main 函数的代码

int main(int argc, char *argv[]) {

/* allow command line calling with arguments -m -b X
where X is a number. order or -m and -b X does not
matter*/
cout << "\Designed By a Student For the Students :)\n";
char choice;
do
{
    cout << "\nEnter the number of variables to be Minimized\n";
    cin >> m;
    if (argc >= 2)
    {
        string arg = argv[1];
        if (arg.find("-m") != -1) {
            show_mid = true;
            if (argc >= 3) {
                arg = argv[2];
                if (arg.find("-b") != -1)
                    MIN_BIT = atoi(argv[3]);
            }
        }
        else if (arg.find("-h") != -1) {
            cout << "-b X\tminimum bits should be X.\n"
                << "-m  \tshow mid process computation.\n"
                << "-h  \tshow this.\n";
                                     return 0;


        }
        else {
            if (arg.find("-b") != -1 && argc >= 3)
                MIN_BIT = atoi(argv[2]);

            if (argc >= 4) {
                arg = argv[3];
                if (arg.find("-m") != -1)
                    show_mid = true;
            }
            else
            {
                cout << "Invalid argument\n"
                    << "-b X\tminimum bits should be X.\n"
                    << "-m  \tshow mid process computation.\n"
                    << "-h  \tshow this.\n";
                                            return 0;

            }
        }
    }

    getinput();
    init();
    cout << "Press 'y' to Reduce Another Expression or 'n' to Close this Application";
    cin >> choice;
} while (choice == 'y');
WINPAUSE;
return 0;

}

正如您在上面看到的，我为此目的使用了一个 do while 循环，但我在这里面临两个问题，即程序在没有用户输入的情况下终止，如果我使用 WINPAUSE，当我按下任意键时程序将退出。递归是答案吗，请提出解决方法。

注意：我使用的是 VS2017 IDE..:)

编辑：新代码

int main(int argc, char *argv[]) {

/* allow command line calling with arguments -m -b X
where X is a number. order or -m and -b X does not
matter*/
cout << "\Designed By a Student For the Students :)\n";
char choice;
    if (argc >= 2)
    {
        string arg = argv[1];
        if (arg.find("-m") != -1) {
            show_mid = true;
            if (argc >= 3) {
                arg = argv[2];
                if (arg.find("-b") != -1)
                    MIN_BIT = atoi(argv[3]);
            }
        }
        else if (arg.find("-h") != -1) {
            cout << "-b X\tminimum bits should be X.\n"
                << "-m  \tshow mid process computation.\n"
                << "-h  \tshow this.\n";


        }
        else {
            if (arg.find("-b") != -1 && argc >= 3)
                MIN_BIT = atoi(argv[2]);

            if (argc >= 4) {
                arg = argv[3];
                if (arg.find("-m") != -1)
                    show_mid = true;
            }
            else
            {
                cout << "Invalid argument\n"
                    << "-b X\tminimum bits should be X.\n"
                    << "-m  \tshow mid process computation.\n"
                    << "-h  \tshow this.\n";

            }
        }
    }
    do
    {
        cout << "\nEnter the number of variables to be Minimized\n";
        cin >> m;
        getinput();
        init();
        cout << "Press 'y' to Reduce Another Expression or 'n' to Close this Application";
        cin >> choice;
    } while (choice == 'y');
WINPAUSE;
return 0;

}

编辑：这是 getinput() 和 init() 的代码

void getinput() {
unsigned in;
int num_bits = 0;
cout << "\nInput value followed by ENTER[^D ends input]\n> ";
while (cin >> in) {
    input_values.push_back(in);
    num_bits = count_bits(in);
    if (num_bits>MIN_BIT)
        MIN_BIT = num_bits;
    cout << "> ";
}
}
/*return min number of bits a number is represented by. used for best output*/
unsigned count_bits(unsigned n) {
    short bit = 0;
    int count = 0;
    while (n>0) {
        bit = n % 2;
        n >>= 1;
        count++;
    }
    return count;
}


void init() {
table.resize(1);
p_group.resize(1);
final_group.resize(1);
create_table();
print_table();
create_p_group();
if (show_mid)
    print_p_group();
create_final_group();
print_final_group();

}

Answer 1

你的逻辑在很多地方都有缺陷，虽然一堆全局变量使它有些模糊，但有些事情是清楚的：

• 在每次迭代时停止重新处理命令行参数。
• 意识到char 的格式化提取不会跳过空格

如果您正确解析命令行参数，则前者可能是正确的。简而言之，如果您发现程序在到达循环之前终止，则意味着您的命令行参数不正确，或者您的解析逻辑已损坏

后者是阻止您的循环正确终止的原因。除非使用整行数据（包括尾随换行符），否则您之前格式化的输入将在输入流上至少留下一个换行符。如果不清除它，为choice 读取的格式化字符将简单地使用它，这显然不是'y' 的值。为了适应这种情况，在读取choice 之前，通过任何换行符丢弃输入流中的任何数据。注意：如果您的输入处理使用像 std::getline 这样的东西，它会消耗（并丢弃）换行符，则不需要此步骤。

你的循环应该是这样的：

int main(int argc, char *argv[])
{
    /* configure command line argument settings ONCE */
    if (argc >= 2)
    {
        string arg = argv[1];
        if (arg.find("-m") != -1) {
            show_mid = true;
            if (argc >= 3) {
                arg = argv[2];
                if (arg.find("-b") != -1)
                    MIN_BIT = atoi(argv[3]);
            }
        }
        else if (arg.find("-h") != -1) {
            cout << "-b X\tminimum bits should be X.\n"
            << "-m  \tshow mid process computation.\n"
            << "-h  \tshow this.\n";
            return 0;

        }
        else {
            if (arg.find("-b") != -1 && argc >= 3)
                MIN_BIT = atoi(argv[2]);

            if (argc >= 4) {
                arg = argv[3];
                if (arg.find("-m") != -1)
                    show_mid = true;
            }
            else
            {
                cout << "Invalid argument\n"
                << "-b X\tminimum bits should be X.\n"
                << "-m  \tshow mid process computation.\n"
                << "-h  \tshow this.\n";
                return 0;
            }
        }
    }

    cout << "Designed By a Student For the Students :)\n";
    char choice = 'y';
    do
    {
        cout << "\nEnter the number of variables to be Minimized\n";
        if (cin >> m)
        {
            getinput();
            init();
            cout << "Press 'y' to Reduce Another Expression or 'n' to Close this Application";

            // flush through eoln, read prompt
            std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
            if (!(std::cin >> choice))
                break;
        }
        else
        {   // could not extract a valid value for m. not much more we can do.
            break;
        }

    } while (choice == 'y');
    return 0;
}

Answer 2

将检查命令行参数的部分移到你的for循环之前，像这样：

cout << "\nEnter the number of variables to be Minimized\n";
cin >> m;
if (argc >= 2)
{
        string arg = argv[1];
        ...
        else
        {
            cout << "Invalid argument\n"
                << "-b X\tminimum bits should be X.\n"
                << "-m  \tshow mid process computation.\n"
                << "-h  \tshow this.\n";
            return 0;
        }
    }
}

do
{
    // your logic, init(), etc.
    cout << "Press 'y' to Reduce Another Expression or 'n' to Close this Application\n";
    cin >> choice;

} while (choice == 'y');

然后是你的

你有很多这样的陈述：

if (arg.find("-m") != -1)

您应该按照std::string::find 中的建议使用string::npos 来测试是否找不到匹配项。所以把它们改成这样：

if (arg.find("-m") != string::npos)

我必须让主函数无效

没有。

不要那样做。 What should main() return in C and C++? int 就是答案。