How should I write an eloquent continue? - c++

Question

So its often come up in my classes that i need to be able to have a program where the user can do things multiple times.
so I write something like
boolean continue=true;
while (continue) {
//do code
boolean isAnswerGood=false;
while (!isAnswerGood) {
cout << "Do you wish to continue" << endl;
string answer;
getline(cin, answer);
if (answer=="y") //or something else
isAnswerGood=true;
else if (answer=="n") {
isAnswerGood=true;
continue=false;
} else
cout << "That wasnt "y" or "n" please type something again" << endl;
}
}
This seems bloated and a lot of code for something so simple. Im willing to think outside of the box with this one, so if anyone can give me a clue as to a better solution to this, id appreciate it.

Answer 1

Break it into separate functions. (Almost always the answer to "how do I write an eloquent...?")
For example:
do {
// something
} while (UserWantsMore());
/////////
bool UserWantsMore() {
std::string answer = GetAnswer("Continue?");
while (!GoodAnswer(answer))
answer = GetAnswer("Please answer 'yes' or 'no'!");
return IsAnswerYes(answer);
}
bool GoodAnswer(const std::string& answer) {
return !answer.empty() && (answer[0] == 'y' || answer[0] == 'n');
}
bool IsAnswerYes(const std::string& answer) {
return !answer.empty() && answer[0] == 'y';
}
std::string GetAnswer(const char* prompt) {
std::cout << prompt << std::cend;
std::string answer;
std::getline(std::cin, answer);
return answer;
}

Answer 2

Sorry, but that is about the right amount of code. C and C++ are relatively verbose compared to some other languages.
However, if this is a common problem, you can abstract it out and make a function out of it. The function would be very similar to what you already have, but then you would call it something like this:
boolean again = true;
while (again) {
// do code
again = ask_user("Do you wish to continue?", "y", "n");
}
The arguments to ask_user() should be the question to ask, the answer to accept that means "yes", and the answer to accept that means "no". The error message (about an unexpected input) can be made using the second and third arguments, so we don't really need to pass it in.
Of course, the problem could be even more complicated than this... what if your code will be used by non-English speakers? If you needed to handle localization, you could make a basic function that has all strings passed in including the error message, and then make a wrapper that the user calls with a language specification. Here's an example, this time using the "infinite loop" in C++:
for (;;) {
// do code
if (!ask_user_yesno("en_us")) // locale: English, USA
break;
}

Answer 3

If you are just wanting to make what you already have do the same thing with a little less code, this will work.
string answer="y";
while (answer=="y") {
//do code
for(;;) {
cout << "Do you wish to continue" << endl;
getline(cin, answer);
if ((answer=="y")||(answer=="n")) break;
cout << "That wasnt \"y\" or \"n\" please type something again" << endl;
}
}
Slightly less code, but slightly more obfuscated:
string answer="y";
while (answer!="n") {
if (answer=="y") {
//do code
} else {
cout << "That wasnt \"y\" or \"n\" please type something again" << endl;
}
cout << "Do you wish to continue" << endl;
getline(cin, answer);
}
Here's a version that uses <termios.h> to get the answer. It uses more code, but behaves more "eloquently".
int getch (void)
{
int c;
struct termios oldt;
struct termios newt;
tcgetattr(STDIN_FILENO, &oldt);
newt = oldt;
newt.c_lflag &= ~ICANON;
tcsetattr(STDIN_FILENO, TCSANOW, &newt);
c = getchar();
tcsetattr(STDIN_FILENO, TCSANOW, &oldt);
return c;
}
bool again (std::string prompt, std::string yes, std::string no)
{
bool doyes = false;
bool dono = false;
for (;;) {
std::cout << prompt;
int c = getch();
std::cout << std::endl;
doyes = (yes.find(c) != yes.npos);
dono = (no.find(c) != no.npos);
if (doyes || dono) break;
std::cout << "Type [" << yes << "] for yes, or [" << no << "] for no.";
std::cout << std::endl;
}
return doyes;
}
You can use it as others have suggested:
do {
// the interesting code
} while (again("Do you wish to continue? ", "y", "n"));

Answer 4

How about something like:
void DoYesCode()
{
// Do the code for the yes stuff...
}
...
do{
cout << "Do you wish to continue" << endl;
string answer;
getline(cin, answer);
if (answer=="y")
DoYesCode();
else if (answer=="n")
break;
else
cout << "That wasnt 'y' or 'n' please type something again" << endl;
} while(true);