C ++ if语句没有返回

Question

int main()
{
char command = 'a';
Monster Goblin;Goblin.HP = 5;Goblin.name = "Goblin";
if(command == 'a'){
    cout<<"At the main Menu, what to do now? Enter H for a list of commands!"<< endl;
    cin>>command;
    switch(command)
    {
        case 'a':
            cout<<"Going to the main menu!"<<endl;
            command = 'a';
            break;
        case 'b':
            cout<<"Going to command line B"<<endl;
            command = 'b';
            break;
        case 'c':
            cout<<"going to command line C"<<endl;
            command = 'c';
            break;
        }

}
if(command == 'b')
{
    cout<<"You made it to command line B"<<endl;
    cout<<"Now lets try to make it go back to the MM!"<<endl;
    command = 'a';
}
if (command == 'c')
{
    cout<<"You made it to command line C"<<endl;
}
}

im trying to get it to when I enter b, it will output going to command line B and the other two lines and then return to the main menu, which is 'a', why is it not returning to the main menu if the command char equals 'a'?

Answer 1

Use a switch :

switch (command) {
case 'a':
    ...
    break;
case 'b':
    ...
    break;
case 'c':
    ...
    break;
default: 
    break;
}

Answer 2

Well there's nothing telling your code to come back at the first condition.

You could do the following:

while(true)
{
if(command == 'a'){
    cout<<"At the main Menu, what to do now? Enter H for a list of commands!"<< endl;
    cin>>command;
    switch(command)
    {
        case 'a':
            cout<<"Going to the main menu!"<<endl;
            command = 'a';
            break;
        case 'b':
            cout<<"Going to command line B"<<endl;
            command = 'b';
            break;
        case 'c':
            cout<<"going to command line C"<<endl;
            command = 'c';
            break;
        }

}
if(command == 'b')
{
    cout<<"You made it to command line B"<<endl;
    cout<<"Now lets try to make it go back to the MM!"<<endl;
    command = 'a';
}
if (command == 'c')
{
    cout<<"You made it to command line C"<<endl;
}
}

Answer 3

If your main menu is just your 'cout' at the top followed by another cin, then there are two solutions.

The solution you may have been thinking of is to wrap your code sample in a while loop.

(while command != "c"){ ... }

This will end the code once you select c. I'm assuming that you don't want to return to the main menu if the player chooses c specifically.

With the current code I'm not exactly thrilled with this method however, as the loop will keep checking the updated command variable. You will have to set the state of command back to 'a' for every command if you don't want to be infinitely looping on the same command. A better solution would be to separate the code into functions.

as an example:

void commandB (){
    cout<<"You made it to command line B"<<endl;
    cout<<"Now lets try to make it go back to the MM!"<<endl;
}

void commandC (){
     cout<<"You made it to command line C"<<endl;
}


int main()
{
    char command = 'a';
    //Monster Goblin;Goblin.HP = 5;Goblin.name = "Goblin";

    while (command != 'c'){
        if(command == 'a'){
            cout<<"At the main Menu, what to do now? Enter H for a list of commands!"<< endl;
            cin>>command;
            switch(command)
            {
                case 'a':
                    cout<<"Going to the main menu!"<<endl;
                    break;
                case 'b':
                    cout<<"Going to command line B"<<endl;
                    commandB(); 
                    command = 'a'; // THIS IS WHAT KEEPS YOU WITHIN THE MM!
                    break;
                case 'c':
                    cout<<"going to command line C"<<endl;
                    commandC();
                    break;
            }

        }
    }
} 

PS : there's no reason to change command to variable 'a' if the variable is already variable 'a':

(your code)

case 'a':
            cout<<"Going to the main menu!"<<endl;
            command = 'a';
            break;

Answer 4

Something like this would probably be considered good and clear programming style:

// Function forward-declarations:
void EnterMainMenu();
void EnterCommandLineB();
void EnterCommandLineC();

int main()
{
    Monster Goblin;
    Goblin.HP = 5;
    Goblin.name = "Goblin";
    EnterMainMenu();
}

void EnterMainMenu()
{
    while(true) // Infinite loop
    {
        char command;
        cout<<"At the main Menu, what to do now? Enter H for a list of commands!"<< endl;
        cin>>command;
        switch(command)
        {
            case 'a':
                cout<<"Going to the main menu!"<<endl;
                // Main menu loop will start again after the next line
                break;
            case 'b':
                cout<<"Going to command line B"<<endl;
                EnterCommandLineB();
                break;
            case 'c':
                cout<<"going to command line C"<<endl;
                EnterCommandLineC();
                break;
        }
    }
}

void EnterCommandLineB()
{
    cout<<"You made it to command line B"<<endl;
    cout<<"Now lets try to make it go back to the MM!"<<endl;
}

void EnterCommandLineC()
{
    cout<<"You made it to command line C"<<endl;
}

Note that after the switch statement the loop will start again from the beginning, including when the EnterCommandLine functions finish execution.