C++ 涉及函数的switch语句的问题

此程序不会提示我输入。我需要帮助正确定义和调用函数。菜单应该显示一个文本字符串，并要求用户选择一个数字1-4，如果用户输入的数字超出该范围，它将要求用户输入一个适当的数字，直到输入1-4为止。menu的值将存储在变量“choice”中，并在开关语句中使用。 在switch语句中，调用了相应的选择函数[getSum、getFactor、getExpo和exit]。它们都需要向用户请求一个整数，并执行一些算术运算，然后向用户返回一些输出文本和计算值。 所有这些都在重复此过程的do while循环内，直到用户选择选项4退出程序，退出函数将向用户返回字符串退出消息，然后程序将终止

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

int menu();
long int getSum();
long int getFactor();
long int getExpo();
string exit();


int main()
{
        const int SUM = 1, // Summation choice
    FACTOR = 2, // Factorial choice
    EXPO = 3, // Exponential choice
    QUIT = 4; // Exit choice


    int choice; // Numerical menu choice
    long int sums, facts, expos;
    string quits;


    // Force the console to print standard notation
    cout << fixed << setprecision(0);

    // Do-While loop controlled by the users choices
    do
    {
        choice = menu();



        switch (choice) // Start switch option
        {
            case SUM: // 1st choice
            sums = getSum();
            break;

        case FACTOR: // 2nd choice
            facts = getFactor();
            break;

        case EXPO: // 3rd choice
            expos = getExpo();
            break;

        case QUIT: // 4th choice
            quits = exit();
            break;

        default: // Default choice
            // Error message for input outside domain
            cout << "Please make a selection of either 1,2,3 or 4.\n\n";
            cin >> choice; // Repeat attempt to gather input from user
        }
    }
    while (menu() != QUIT);

    return 0;
}

int menu()
{
    int choice;
    cout << "\n\t\tMathematical Menu\n" // Header
    << "1) Summation\n" // 1st choice
    << "2) Factorial\n" // 2nd choice
    << "3) Exponential\n" // 3rd choice
    << "4) Exit Program\n\n" // 4th choice
    << "Please make a selection\n" // Ask user for imput choice
    << "of either 1,2,3 or 4.\n\n";
    cin >> choice; // Gather input from user
    return choice;
}

long int getSum()
{
    int total = 0, userNum, counter;

    // Ouput statement to user
    cout << "Please enter a positive integer value greater than 0 \n"
    << "and less than 10,000,000.\n\n";
    cin >> userNum; // Repeat attempt to gather input from user

    // Compare input to domain
    if (userNum < 0 || userNum > 10000000)
    {
        // Error message for input outside domain
        cout << "Please check your entry and try again.\n\n";
    cin >> userNum; // Repeat attempt to gather input from user
}

// Perform arithmetic summation
for (counter = 1; counter <= userNum; counter++)
{
    total += counter; // Running count
}
cout << "The total value for the added numbers 1 to \n"
<< userNum << " is:\n"<<total;
return total;
}

long int getFactor()
{
int total, userNum, counter;

total = 1;
// Output statement to user
cout << "Please enter a positive integer from 0 \n"
<< "and less than 100.\n\n";
cin >> userNum; // Gather input from user

// Compare input to domain
if (userNum > 100 || userNum < 0)
{
    // Error message if input is outside domain
    cout << "Please check your entry and try again.\n\n";
    cin >> userNum; // Repeat attempt to gather input from user
}

// Perform arithmetic factorial
for (counter = 1; counter <= userNum; counter++)
{
    total *= counter; // Running count
}

// Display arithmetic output to user
cout << "The total value for the multiplied numbers 1 to \n"
<< userNum << " is:\n";
return total;
}

long int getExpo()
{
int total, userNum, counter;

total = 0;
// Output statement to user
cout << "Please enter a positive integer from 0 \n"
<< "and less than 100.\n\n";
cin >> userNum; // Gather input from user

// Compare input to domain
if (userNum > 100 || userNum < 0)
{
    // Error message if input is outside domain
    cout << "Please check your entry and try again.\n\n";
    cin >> userNum; // Repeat attempt to gather input from user
}

// Perform arithmetic exponential
for (counter = 1; counter <= userNum; counter++)
{
    total = pow(2.0, userNum); // Running count
}

// Display arithmetic output to user
cout << "The total value for the exponential function is \n";
return total;
}

string exit()
{
// Exit message
return "Don't be gone for too long...\n";

}`

不要多次调用菜单：它将提示用户两次

无论如何，不要从完成所有工作的函数返回任何内容。成功

void getExpo(){ ...; return; }

电话：

---

你需要更好地解释你的问题。你是在要求别人向你解释你的代码在做什么吗？你得到的结果不正确吗？a问题是什么？b代码示例太长，请提供与您的问题匹配的最短代码。c这不是解决你的家庭作业的网站。getSum似乎少了几行？@Flovdis我没有要求任何人解决我的“家庭作业”。显然，我已经尝试自己解决这个问题，并试图清除任何编译器错误。我只是不明白如何用我定义的函数恰当地调用和传递值。@怀疑这只是个bug，很容易看到-请参阅我的答案和注释。我认为有足够的斗争需要帮助。我不相信我给菜单打了不止一次电话。我有一个原型，一个电话，然后是一个定义。感谢您的尝试，但是我尝试使用您的建议进行修复，仍然得到0个输入数据的提示，等等。您必须执行{choice=menu；…和…}while menu！=退出，1+1=2。只需尝试该程序：除了在第一次选择后提示两次，即只有一次出现其他错误时才会提示。对于“1”，考虑到用户输入的上限，32位int是不够的。对于“2”，结果不会显示，使用100作为上限不会在任何整数模式下产生有意义的结果。我没有检查“3”。谢谢@luane，我的IDE一直有问题。我以为我写的所有代码都不管用。我创建了一个新项目，从你的答案中吸取了一些技巧，我很乐意去做。再次感谢你。

getExpo();