如何在迭代后重复从主函数重复执行

所以我有一个程序，可以简化布尔表达式。我要实现的是，在简化第一个表达式的结束时，我希望用户选择他是否要简化另一个表达式或仅退出控制台应用程序（程序）。

这是主函数的代码

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 Minimizedn";
    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 Xtminimum 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 argumentn"
                    << "-b Xtminimum 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;

}

如上所以看到，我在循环中使用了do do n do to to to the，我在这里遇到了两个问题，即程序终止而无需接受用户输入，如果我使用winpause，当我按任何键时，程序将退出。是递归的答案，请建议解决方法。

注意：我使用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 Xtminimum 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 argumentn"
                    << "-b Xtminimum bits should be X.n"
                    << "-m  tshow mid process computation.n"
                    << "-h  tshow this.n";
            }
        }
    }
    do
    {
        cout << "nEnter the number of variables to be Minimizedn";
        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（）

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();

}