我是全新的，我不知道如何问这个或者甚至搜索什么。
案例是这样的：我想浏览带有几个子菜单的菜单。在这个例子中，我将只使用“选项”和“游戏”来说明我的意思。假设您有一个包含3个选项的菜单。
1 - 开始
2 - 选项
3 - 退出
选择选项会带您进入另一个菜单。那会是什么样的
1 - 难度
2 - 声音
3 - 返回
根据您从这里的位置，显然会有更多的子菜单。 我试过嵌套do-while循环和各种各样的东西，但我只是没有足够的理解知道它是什么我做错了。
这是我到目前为止所做的：
#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int choice;
do{
cout << "Main Menu\n";
cout << "Please make your selection\n";
cout << "1 - Start game\n";
cout << "2 - Options\n";
cout << "3 - Quit\n";
cout << "Selection: ";
cin >> choice;
switch(choice) {
case 1:
cout << "Pew pew!\n";
break;
case 2:
cout <<"????\n";
break;
case 3:
cout << "Goodbye!";
break;
default:
cout << "Main Menu\n";
cout << "Please make your selection\n";
cout << "1 - Start game\n";
cout << "2 - Options\n";
cout << "3 - Quit\n";
cout << "Selection: ";
cin >> choice;
}
} while(choice !=3);
system("PAUSE");
return EXIT_SUCCESS;
}
其功能类似于常规菜单。但我不知道从哪里开始。我查阅过一些书籍，但发现与此有关的任何东西都是随机的。任何帮助或示例将不胜感激。
嵌套大量循环后发生的事情只是使每次循环同时执行。我如何防止这种情况发生？做出更多选择？ （选择1-2-3等？还是什么？）
答案 0 :(得分：4)
好的伙计们。谢谢你的帮助。这就是我最终得到的结果。 它按照我的要求运行，并且通过max_的例子和Mike B的评论我觉得这很有效。
非常感谢大家=）
#include <iostream>
#include <cstdlib>
using namespace std;
void menu();
void mainMenu();
void optionsMenu();
void options();
int choice1 = 0;
int choice2 = 3;
int main(int argc, char** argv) {
menu();
return 0;
}
void menu(){
do {
choice2 = 0;
mainMenu();
switch(choice1) {
case 1:
cout << "Pew pew!\n";
break;
case 2:
options();
break;
case 3:
break;
}
} while(choice1 != 3);
}
void options(void) {
do {
optionsMenu();
switch(choice2){
case 1:
cout << "So difficult!\n";
break;
case 2:
cout << "Beep!\n";
break;
case 3:
break;
default:
break;
}
} while(choice2 != 3);
}
void mainMenu(void) {
cout << "Main Menu\n";
cout << "1 - Start game\n";
cout << "2 - Options\n";
cout << "3 - Quit\n";
cout << "Please choose: ";
cin >> choice1;
}
void optionsMenu(void) {
cout << "Options Menu\n";
cout << "1 - Difficulty\n";
cout << "2 - Sound";
cout << "3 - Back\n";
cout << "Please choose: ";
cin >> choice2;
}
答案 1 :(得分：2)
我建议您在这里更改一些内容。您熟悉面向对象的设计吗？如果没有，强烈建议你阅读一下如果你想用C ++编写代码（或者只是编写代码，因为它是许多编程语言的一个重要方面）< / p>
考虑将每个菜单和子菜单视为单个对象。每次进入循环时，使用对象指针调用打印当前菜单文本的方法。
然后，正常地从用户那里获取输入，并更改您现在正在使用的菜单对象。
这可能不是执行控制台菜单的最理想方式，但它将为您提供一个非常强大的基础，以确定面向对象的编程是如何工作的。
我附上了一个例子：
#include <iostream>
#include <string>
class BaseMenu
{
public:
BaseMenu() { m_MenuText = "This shouldn't ever be shown!"; } // This is the constructor - we use it to set class-specific information. Here, each menu object has its own menu text.
virtual ~BaseMenu() { } // This is the virtual destructor. It must be made virtual, else you get memory leaks - it's not a quick explaination, I recommend you read up on it
virtual BaseMenu *getNextMenu(int iChoice, bool& iIsQuitOptionSelected) = 0; // This is a 'pure virtual method', as shown by the "= 0". It means it doesn't do anything. It's used to set up the framework
virtual void printText() // This is made virtual, but doesn't *have* to be redefined. In the current code I have written, it is not redefined as we store the menu text as a string in the object
{
std::cout << m_MenuText << std::endl;
}
protected:
std::string m_MenuText; // This string will be shared by all children (i.e. derived) classes
};
class FirstMenu : public BaseMenu // We're saying that this FirstMenu class is a type of BaseMenu
{
FirstMenu()
{
m_MenuText = "Main Menu\n" // What we are doing here is setting up the string to be displayed later
+ "Please make your selection\n" // What we are doing here is setting up the string to be displayed later
+ "1 - Start game\n" // What we are doing here is setting up the string to be displayed later
+ "2 - Options\n" // What we are doing here is setting up the string to be displayed later
+ "3 - Quit\n" // What we are doing here is setting up the string to be displayed later
+ "Selection: "; // What we are doing here is setting up the string to be displayed later
}
BaseMenu *getNextMenu(int choice, bool& iIsQuitOptionSelected) // This is us actually defining the pure virtual method above
{
BaseMenu *aNewMenu = 0; // We're setting up the pointer here, but makin sure it's null (0)
switch (choice) // Notice - I have only done "options". You would obviously need to do this for all of your menus
{
case 2:
{
aNewMenu = new SecondMenu; // We're creating our new menu object here, and will send it back to the main function below
}
case 3:
{
// Ah, they selected quit! Update the bool we got as input
iIsQuitOptionSelected = true;
}
default:
{
// Do nothing - we won't change the menu
}
}
return aNewMenu; // Sending it back to the main function
}
};
class SecondMenu : public BaseMenu
{
SecondMenu()
{
m_MenuText = "OptionsMenu\n"
+ "Please make your selection\n"
+ "1 - ????"
+ "2 - dafuq?";
}
BaseMenu *getNextMenu(int choice, bool& iIsQuitOptionSelected) // This is us actually defining the pure virtual method above
{
BaseMenu *aNewMenu = 0; // We're setting up the pointer here, but makin sure it's null (0)
switch (choice) // Notice - I have only done options. You would obviously need to do this for all of your menus
{
case 1:
{
aNewMenu = new FirstMenu; // We're creating our new menu object here, and will send it back to the main function below
}
break;
case 2:
{
aNewMenu = new FirstMenu; // We're creating our new menu object here, and will send it back to the main function below
}
break;
default:
{
// Do nothing - we won't change the menu
}
}
return aNewMenu; // Sending it back to the main function
}
};
int main (int argc, char **argv)
{
BaseMenu* aCurrentMenu = new FirstMenu; // We have a pointer to our menu. We're using a pointer so we can change the menu seamlessly.
bool isQuitOptionSelected = false;
while (!isQuitOptionSelected) // We're saying that, as long as the quit option wasn't selected, we keep running
{
aCurrentMenu.printText(); // This will call the method of whichever MenuObject we're using, and print the text we want to display
int choice = 0; // Always initialise variables, unless you're 100% sure you don't want to.
cin >> choice;
BaseMenu* aNewMenuPointer = aBaseMenu.getNextMenu(choice, isQuitOptionSelected); // This will return a new object, of the type of the new menu we want. Also checks if quit was selected
if (aNewMenuPointer) // This is why we set the pointer to 0 when we were creating the new menu - if it's 0, we didn't create a new menu, so we will stick with the old one
{
delete aCurrentMenu; // We're doing this to clean up the old menu, and not leak memory.
aCurrentMenu = aNewMenuPointer; // We're updating the 'current menu' with the new menu we just created
}
}
return true;
}
请注意，这可能对于开始而言有点复杂。我强烈建议您阅读其他人发布的答案。它应该为您提供一些如何做的方法，您可以从基础进展到更复杂，检查每个变化。
答案 2 :(得分：1)
看看你想要做什么，我会改变你如何确保用户仍然希望首先玩游戏。看一下使用while循环来检查一个变量是真还是假（人们倾向于使用布尔变量（bool）），设置为1或0的int将执行相同的操作。这消除了对do-while的需要。建议读取控制逻辑（if / else，while，for循环）和逻辑运算符（＆amp;＆amp; - and，|| - 或，！= - 不等于）。控制逻辑使您的代码执行不同的操作，布尔值可以快速检查是/否场景，逻辑运算符允许您在一个if语句中检查多个项目。 一些阅读：Loops
编辑：有更多的阅读材料链接，没有代表发布它们。
其次，使用另一个变量（int或任何适合您的变量）来跟踪您所在的屏幕。 根据此选择，显示不同的选项，但仍然使用输入1,2,3来决定下一个操作。
在一些可怕的伪代码中，这是我倾向于：
main()
{
int choice
int screen = 1
bool running = true
while(running) {
//Screen 1, Main menu
if(screen == 1) {
cout << stuff
cout << stuff
cout << option 1
cout << option 2
cout << option 3
cout << selection:
cin >> choice
}
else if(screen == 2){
//options screen here
}
else {
//default/error message
}
//add some choice logic here
if(screen == 1 && choice == 3){
//being on screen one AND choice three is quit
running = false;
}
else if(screen == 1 && choice == 2){
//etc..
}
}
}
这是我的第一个正确答案，所有可怕的批评都得到了很好的回应。
答案 3 :(得分：1)
这个怎么样（dunno，如果它编译虽然）：
#include <cstdlib>
#include <iostream>
using namespace std;
int GetInput()
{
int choice;
cin >> choice;
return choice;
}
void DisplayMainMenu()
{
cout << "Main Menu\n";
cout << "Please make your selection\n";
cout << "1 - Start game\n";
cout << "2 - Options\n";
cout << "3 - Quit\n";
cout << "Selection: ";
}
void DisplayOptionsMenu()
{
cout << "Options Menu\n";
cout << "Please make your selection\n";
cout << "1 - Difficulty\n";
cout << "2 - Sound\n";
cout << "3 - Back\n";
cout << "Selection: ";
}
void Options()
{
int choice = 0;
do
{
system("cls");
DisplayOptionsMenu();
choice = GetInput();
switch(choice)
{
case 1:
cout << "difficulty stuff";
break;
case 2:
cout << "sound stuff";
break;
case 3:
break;
default:
break;
}
} while(choice!=3);
}
int main(int argc, char *argv[])
{
int choice = 0;
do
{
system("cls");
DisplayMainMenu();
choice = GetInput();
switch(choice) {
case 1:
cout << "Pew pew!\n";
break;
case 2:
Options();
break;
case 3:
cout << "Goodbye!";
break;
default:
break;
}
} while(choice!=3);
system("PAUSE");
return EXIT_SUCCESS;
}