my c++ contains a conditional loop but its producing a infinite loop

[LIQUIDFOX00200]

please take a look at this code:-

 

 

#include <iostream>
#include <string>

using namespace std;

void PrintIntro();
void PlayGame();
string GetGuess();
bool AskToPlayAgain();

// the entry point for our application
int main()
{ 
    bool bPlayAgain = false;
    do {
    PrintIntro();
    PlayGame();
    AskToPlayAgain();
    bPlayAgain = AskToPlayAgain;
    } 
    while (bPlayAgain);

    return 0; // exit the application
}

// introduce the game
void PrintIntro()
{
    constexpr int WORLD_LENGTH = 9;
    cout << "Welcome to Bulls and Cows, a fun word puzzel game.\n";
    cout << "Can you guess the " << WORLD_LENGTH;
    cout << " letter isogram I'm thinking of?\n";
    cout << endl;
    return;
}

void PlayGame()
{
    // loop for the number of turns asking for guesses
    constexpr int NUMBER_OF_TURNS = 5;
    for (int count = 1; count <= NUMBER_OF_TURNS; count++) {
        string Guess = GetGuess();
        cout << "Your guess was: " << Guess << endl;
        cout << endl;
    }
}

string GetGuess()
{
    // get a guess from the player
    cout << "Enter your guess: ";
    string Guess = "";
    getline(cin, Guess);
    return Guess;
}

bool AskToPlayAgain()
{ //Ask the player that if he/she wants to play again or not! 
    cout << "Do you wanna play again? (Y/N)";
    string Response = "";
    getline(cin, Response);
    return (Response[0] == 'y') || (Response[0] == 'Y');
}
 

 

 

 

 

as you can see i programmed a conditional loop (response == 'y') but its producing a infinite loop!!! whatever i press it starts again!!! 

 

 

please help

[Ithanul]

I think your issue is with the bool AskToPlayAgain function.  Most likely what it is returning.

 

Ah, ok, reading around a bit:  0 = false, 1 = true.

[T3500]

I'm not a programmer but my brother says:

 

askToPlayAgain is never declared as a variable is it?

 

"

AskToPlayAgain();

bPlayAgain = AskToPlayAgain();

"

 

just needs to be

 

"

bPlayAgain = AskToPlayAgain();

"

[Unimportant]

@LIQUIDFOX00200

 

First of all, as already mentioned, you probably want to call AskToPlayAgain only once, but the second call, missing the () results in different behavior:

    AskToPlayAgain();
    bPlayAgain = AskToPlayAgain;  //<== AskToPlayAgain, without (), does not call the function but decays into the function's memory address.
				  //Since the memory address will never be 0 it will always result in bPlayAgain being set to true.

 

[LIQUIDFOX00200]

11 hours ago, LIQUIDFOX00200 said:

bPlayAgain = AskToPlayAgain;

So I missed () in asktoplayagain ?? 

11 hours ago, Ithanul said:

I think your issue is with the bool AskToPlayAgain function.  Most likely what it is returning.

 

Ah, ok, reading around a bit:  0 = false, 1 = true.

 

9 hours ago, Unimportant said:

@LIQUIDFOX00200

 

First of all, as already mentioned, you probably want to call AskToPlayAgain only once, but the second call, missing the () results in different behavior:

    AskToPlayAgain();
    bPlayAgain = AskToPlayAgain;  //<== AskToPlayAgain, without (), does not call the function but decays into the function's memory address.
				  //Since the memory address will never be 0 it will always result in bPlayAgain being set to true.

 

 

10 hours ago, T3500 said:

I'm not a programmer but my brother says:

 

askToPlayAgain is never declared as a variable is it?

 

"

AskToPlayAgain();

bPlayAgain = AskToPlayAgain();

"

 

just needs to be

 

"

bPlayAgain = AskToPlayAgain();

"

 

11 hours ago, Ithanul said:

I think your issue is with the bool AskToPlayAgain function.  Most likely what it is returning.

 

Ah, ok, reading around a bit:  0 = false, 1 = true.

 

[LIQUIDFOX00200]

9 hours ago, Unimportant said:

@LIQUIDFOX00200

 

First of all, as already mentioned, you probably want to call AskToPlayAgain only once, but the second call, missing the () results in different behavior:

    AskToPlayAgain();
    bPlayAgain = AskToPlayAgain;  //<== AskToPlayAgain, without (), does not call the function but decays into the function's memory address.
				  //Since the memory address will never be 0 it will always result in bPlayAgain being set to true.

 

okay its fixed but now its asking 2 times!! that if you want to play again or not!!! first choice is ignored and second choice will work!

 

[HunterAP]

3 hours ago, LIQUIDFOX00200 said:

okay its fixed but now its asking 2 times!! that if you want to play again or not!!! first choice is ignored and second choice will work!

 

Just change:

AskToPlayAgain();
bPlayAgain = AskToPlayAgain;

To:

bPlayAgain = AskToPlayAgain();

When you run the first line in the first snippet of code, the return value of the function is not stored anywhere, and it's just thrown away.

When you run the second line of that same snippet, you're calling the memory address of the function, which for all intents and purposes of this code is simply running the function again.

 

Using parenthesis for function calls is a basic and fundamental part of basically all C based programming. Make sure not to do something like this is future programs: always remember to use parenthesis when you call functions.

 

The second code snippet that is just one line fixes your problem: it only calls the function twice, and makes sure that it is stored in a variable that you can use later, and it has the parenthesis in it to prevent issues with calling the function.

 

The final main function should look like this:

int main()
{ 
    bool bPlayAgain = false;
    do {
    PrintIntro();
    PlayGame();
    bPlayAgain = AskToPlayAgain();
    } 
    while (bPlayAgain);

    return 0; // exit the application
}

 

[LIQUIDFOX00200]

1 hour ago, HunterAP said:

Just change:

AskToPlayAgain();
bPlayAgain = AskToPlayAgain;

To:

bPlayAgain = AskToPlayAgain();

When you run the first line in the first snippet of code, the return value of the function is not stored anywhere, and it's just thrown away.

When you run the second line of that same snippet, you're calling the memory address of the function, which for all intents and purposes of this code is simply running the function again.

 

Using parenthesis for function calls is a basic and fundamental part of basically all C based programming. Make sure not to do something like this is future programs: always remember to use parenthesis when you call functions.

 

The second code snippet that is just one line fixes your problem: it only calls the function twice, and makes sure that it is stored in a variable that you can use later, and it has the parenthesis in it to prevent issues with calling the function.

 

The final main function should look like this:

int main()
{ 
    bool bPlayAgain = false;
    do {
    PrintIntro();
    PlayGame();
    bPlayAgain = AskToPlayAgain();
    } 
    while (bPlayAgain);

    return 0; // exit the application
}

 

 

15 hours ago, Unimportant said:

@LIQUIDFOX00200

 

First of all, as already mentioned, you probably want to call AskToPlayAgain only once, but the second call, missing the () results in different behavior:

    AskToPlayAgain();
    bPlayAgain = AskToPlayAgain;  //<== AskToPlayAgain, without (), does not call the function but decays into the function's memory address.
				  //Since the memory address will never be 0 it will always result in bPlayAgain being set to true.

 

 

[Unimportant]

3 hours ago, HunterAP said:

 

AskToPlayAgain();
bPlayAgain = AskToPlayAgain;

 

When you run the second line of that same snippet, you're calling the memory address of the function, which for all intents and purposes of this code is simply running the function again.

No, it doesn't. A function name without the parentheses decays into a pointer to the function's memory address. It does not call/execute the function:

bPlayAgain = AskToPlayAgain;
bPlayAgain = &AskToPlayAgain;  //Both these lines are equivalent.

So, all that line does is assign the memory address of the AskToPlayAgain function to the bPlayAgain bool variable, using normal conversion rules, which will always result in bPlayAgain being set to true.