c++ - 到达 void 函数的结尾会产生意想不到的结果

Question

我写了一个简单的程序来管理姓名列表(下面是程序的一部分)。我希望函数“choice()”结束并返回到 main()——从而结束程序——当用户对变量“option”的输入为 4 时。然而，choice() 往往会重复几次即使在将 4 分配给“选项”之后，例如:

Would you like to:    (1) - Add name(s) to the list    (2) - Remove name(s) from the list    (3) - View the list    (4) - Exit the program(type 1, 2, 3, or 4): 4Exiting Program...Would you like to:    (1) - Add name(s) to the list    (2) - Remove name(s) from the list    (3) - View the list    (4) - Exit the program(type 1, 2, 3, or 4): 4Exiting Program...ryan$ _

In the absence of the entire program, what do you think may be causing the problem? Sorry if there is not enough information to make a recommendation.

You can see I commented "exit(0)" from the end of the fourth if statement of choice()--in which choice() should reach its end and return to main(). I would prefer to avoid this statement but it is the only solution I can make work at the moment. Thanks for your help.

-Ryan

int main()
{
     cout << endl 
          << "WELCOME TO THE NAME LIST PROGRAM" << endl
          << "--------------------------------" << endl;

    choice();

    return 0;
}


void choice()
{
    string option = "0";

    cout << "\nWould you like to:"             << endl
     << "\t(1) - Add name(s) to the list"      << endl
     << "\t(2) - Remove name(s) from the list" << endl
     << "\t(3) - View the list"                << endl
     << "\t(4) - Exit the program"             << endl
     << "(type 1, 2, 3, or 4): ";

    getline(cin,option);

    while( option != "1" && option != "2" &&
           option != "3" && option != "4"  )
    {
        cout << "\nInvalid input, reenter: ";
        getline(cin,option);
    }

    if(option == "1")
    {
        appendNames();
        choice();
    }
    else if(option == "2")
    {
        string flag = "menu";
        removeNames(flag);
        choice();
    }

    else if(option == "3")
    {
        viewNames();
        choice();
    }

    else if(option == "4")
    {
        cout << "\n\nExiting Program...\n\n";
        // exit(0);
    }
}

Answer 1

我认为问题来自对函数工作方式的误解。假设您有一个显示您姓名的函数:

void showName()
{
    cout << "Ryan" << endl;
}

现在假设您想重复 10 次。在您特定的编程方式中，您可能会做如下事情:

void showName(int times)
{
    cout << "Ryan" << endl;

    if ( times > 0 ) {
        showName( --times );
    }

    return;
}

int main()
{
    showName( 10 );
}

这种编程方式称为递归，但我很确定您是偶然接触到的。递归在某些情况下很方便，但它不是做简单事情的常用方法，比如重复某事。让我们看看调用该函数 10 次的实际程序:

void showName()
{
    cout << "Ryan" << endl;
}

int main()
{
    for(int i = 0; i < 10; ++i) {
        showName();
    }
}

每次调用函数 showName() 时，被调用方都必须返回给调用方。如果您执行示例中显示的操作:

if( option == "1" )
{
    appendNames();
    choice();
}

然后，您实际上调用了 choice() 的次数与用户选择除 4(退出)之外的任何其他选项的次数相同。这解释了为什么您必须调用 exit( 0 ) 才能立即生效。

总而言之，您的程序应该是:

int main()
{
     cout << endl 
          << "WELCOME TO THE NAME LIST PROGRAM" << endl
          << "--------------------------------" << endl;

    while( !choice() );

    return 0;
}


bool choice()
{
    bool toret = false;
    string option = "0";

    cout << "\nWould you like to:"             << endl
     << "\t(1) - Add name(s) to the list"      << endl
     << "\t(2) - Remove name(s) from the list" << endl
     << "\t(3) - View the list"                << endl
     << "\t(4) - Exit the program"             << endl
     << "(type 1, 2, 3, or 4): ";

    getline(cin,option);

    while( option != "1" && option != "2" &&
           option != "3" && option != "4"  )
    {
        cout << "\nInvalid input, reenter: ";
        getline(cin,option);
    }

    if(option == "1")
    {
        appendNames();
    }
    else if(option == "2")
    {
        removeNames();
    }

    else if(option == "3")
    {
        viewNames();
    }

    else if(option == "4")
    {
        cout << "\n\nExiting Program...\n\n";
        toret = true;
    }

    return toret;
}

希望这对您有所帮助。