I tried to make a an introduction to a "game", and in its functions I made some Yes/No, 1/2/3, situations.
Im new to this however it wasn't that difficult, worked perfectly. The problem appeared when handling with invalid inputs. So this is what the code looks like by now:
#include "Introduction.h"
#include "GameConstants.h"
#include "PlayerCharacter.h"
#include <iostream>
#include <windows.h>
using namespace std;
Introduction::Introduction()
{
}
/////////Function N.1///////////
void Introduction::presentation()
{
char confirm;
string enteredName;
cout << constants.line() << "Welcome traveler! What is the name?" << endl;
getline(cin,enteredName);// Gets the WHOLE LINE as the name.
while (confirm != 'Y') //If the player doesn't confirm the name with 'Y' in will run again until it does.
{
cout << constants.xline() << "Your name is " << enteredName << " right? (Y/N)" << endl;
cin >> confirm; //The player's answer
cin.sync(); //Only takes the first character
confirm = toupper(confirm); //Turns player message into CAPS for easier detection in the "if" statements
if (confirm == 'N'){ //If not the correct name, gives another chance
cout << constants.xline() << "Please, tell me your name again..." << endl;
cin >> enteredName;
cin.sync();}
if ((confirm != 'Y')&&(confirm != 'N')){ //If an invalid input is entered, gives another chance. And insults you.
cout << constants.xline() << "Fool Go ahead, just enter your name again." << endl;
cin >> enteredName;
cin.sync();}
}
if (confirm == 'Y'){ //When the answer is yes ('Y') /* Uneeded line */
PC.setName(enteredName); //Saves the name
cout << constants.xline() << "Excellent! I have a few more questions for you " << PC.name() << "..." << endl;
}
}
//////////Function N.2///////////
void Introduction::difSelection(){
int selectedDif = 0; //Variable to store selected difficulty whitin this function.
Sleep(2500);
cout << constants.xline() << "What kind of adventure do you want to take part in?" << endl;
Sleep(2500); //Wait 2,5 s
cout << "\n1= Easy\n2= Normal\n3= Hard" << endl;
while(selectedDif != 1&&2&&3){ //Selected option must be 1/2/3 or will run again
cin >> selectedDif; //Sets the user selected difficulty
cin.sync(); //Gets only first character
if((selectedDif != 1||2||3)&&(!(selectedDif))){ //If the input isn't 1/2/3 AND is an invalid character, this will run. And it'll start again
cout << constants.xline() << "Criminal scum. Go again." << endl;
cin.clear();
cin.ignore();
}
if(selectedDif != 1&&2&&3){ //If selected option isn't 1/2/3, this will run and will loop again. However I know this conflicts with the previous statement since this will run anyways.
cout << constants.xline() << "Wrong input, please try again." << endl;
}
else if(selectedDif == 1){
constants.setDiff(1);
constants.setStatPoints(15);
} else if(selectedDif == 2){
constants.setDiff(2);
constants.setStatPoints(10);
} else if (selectedDif == 3){
constants.setDiff(3);
constants.setStatPoints(5);}
}
}
The first function works perfectly you can type "aaa" or "a a a" and will work. However I'd like to know if there's a simpler way to do it. (Understandable for a beginner, just started 3 days ago lol; if it includes some advanced or less known code prefer to stay like this by now).
Now, the second one, I really have no idea how to fix it. I need something that if the user's input was an invalid character type, throw certain message, and if it's an int type, but out of the range, another message. And of course, run again if it fails. Did a lot of search and couldn't find anything that meet this requirements.
To check if the user input is an int, you could use the good() function.
int val;
cin >> val;
if( cin.good() ) {
// user input was a valid int
} else {
// otherwise
}
As for the range check, the syntax is a bit different.
This returns true if the number is not equal to 1 nor 2 nor 3:
selectedDif != 1 && selectedDif != 2 && selectedDif != 3
Another shorter way would be to use:
selectedDif < 1 || selectedDif > 3
Another thing, in c++, there are two keywords break and continue which will allow to reduce the code in the loops.
Related
Please note that I am a complete beginner at C++. I'm trying to write a simple program for an ATM and I have to account for all errors. User may use only integers for input so I need to check if input value is indeed an integer, and my program (this one is shortened) works for the most part.
The problem arises when I try to input a string value instead of an integer while choosing an operation. It works with invalid value integers, but with strings it creates an infinite loop until it eventually stops (unless I add system("cls"), then it doesn't even stop), when it should output the same result as it does for invalid integers:
Invalid choice of operation.
Please select an operation:
1 - Balance inquiry
7 - Return card
Enter your choice and press return:
Here is my code:
#include <iostream>
#include <string>
using namespace std;
bool isNumber(string s) //function to determine if input value is int
{
for (int i = 0; i < s.length(); i++)
if (isdigit(s[i]) == false)
return false;
return true;
}
int ReturnCard() //function to determine whether to continue running or end program
{
string rtrn;
cout << "\nDo you wish to continue? \n1 - Yes \n2 - No, return card" << endl;
cin >> rtrn;
if (rtrn == "1" and isNumber(rtrn)) { return false; }
else if (rtrn == "2" and isNumber(rtrn)) { return true; }
else {cout << "Invalid choice." << endl; ReturnCard(); };
return 0;
}
int menu() //function for operation choice and execution
{
int choice;
do
{
cout << "\nPlease select an operation:\n" << endl
<< " 1 - Balance inquiry\n"
<< " 7 - Return card\n"
<< "\nEnter your choice and press return: ";
int balance = 512;
cin >> choice;
if (choice == 1 and isNumber(to_string(choice))) { cout << "Your balance is $" << balance; "\n\n"; }
else if (choice == 7 and isNumber(to_string(choice))) { cout << "Please wait...\nHave a good day." << endl; return 0; }
else { cout << "Invalid choice of operation."; menu(); }
} while (ReturnCard()==false);
cout << "Please wait...\nHave a good day." << endl;
return 0;
}
int main()
{
string choice;
cout << "Insert debit card to get started." << endl;
menu();
return 0;
}
I've tried every possible solution I know, but nothing seems to work.
***There is a different bug, which is that when I get to the "Do you wish to continue?" part and input any invalid value and follow it up with 2 (which is supposed to end the program) after it asks again, it outputs the result for 1 (continue running - menu etc.). I have already emailed my teacher about this and this is not my main question, but I would appreciate any help.
Thank you!
There are a few things mixed up in your code. Always try to compile your code with maximum warnings turned on, e.g., for GCC add at least the -Wall flag.
Then your compiler would warn you of some of the mistakes you made.
First, it seems like you are confusing string choice and int choice. Two different variables in different scopes. The string one is unused and completely redundant. You can delete it and nothing will change.
In menu, you say cin >> choice;, where choice is of type int. The stream operator >> works like this: It will try to read as many characters as it can, such that the characters match the requested type. So this will only read ints.
Then you convert your valid int into a string and call isNumber() - which will alway return true.
So if you wish to read any line of text and handle it, you can use getline():
string inp;
std::getline(std::cin, inp);
if (!isNumber(inp)) {
std::cout << "ERROR\n";
return 1;
}
int choice = std::stoi(inp); // May throw an exception if invalid range
See stoi
Your isNumber() implementation could look like this:
#include <algorithm>
bool is_number(const string &inp) {
return std::all_of(inp.cbegin(), inp.cend(),
[](unsigned char c){ return std::isdigit(c); });
}
If you are into that functional style, like I am ;)
EDIT:
Btw., another bug which the compiler warns about: cout << "Your balance is $" << balance; "\n\n"; - the newlines are separated by ;, so it's a new statement and this does nothing. You probably wanted the << operator instead.
Recursive call bug:
In { cout << "Invalid choice of operation."; menu(); } and same for ReturnCard(), the function calls itself (recursion).
This is not at all what you want! This will start the function over, but once that call has ended, you continue where that call happened.
What you want in menu() is to start the loop over. You can do that with the continue keyword.
You want the same for ReturnCard(). But you need a loop there.
And now, that I read that code, you don't even need to convert the input to an integer. All you do is compare it. So you can simply do:
string inp;
std::getline(std::cin, inp);
if (inp == "1" || inp == "2") {
// good
} else {
// Invalid
}
Unless that is part of your task.
It is always good to save console input in a string variable instead of another
type, e.g. int or double. This avoids trouble with input errors, e.g. if
characters instead of numbers are given by the program user. Afterwards the
string variable could by analyzed for further actions.
Therefore I changed the type of choice from int to string and adopted the
downstream code to it.
Please try the following program and consider my adaptations which are
written as comments starting with tag //CKE:. Thanks.
#include <iostream>
#include <string>
using namespace std;
bool isNumber(const string& s) //function to determine if input value is int
{
for (size_t i = 0; i < s.length(); i++) //CKE: keep same variable type, e.g. unsigned
if (isdigit(s[i]) == false)
return false;
return true;
}
bool ReturnCard() //function to determine whether to continue running or end program
{
string rtrn;
cout << "\nDo you wish to continue? \n1 - Yes \n2 - No, return card" << endl;
cin >> rtrn;
if (rtrn == "1" and isNumber(rtrn)) { return false; }
if (rtrn == "2" and isNumber(rtrn)) { return true; } //CKE: remove redundant else
cout << "Invalid choice." << endl; ReturnCard(); //CKE: remove redundant else + semicolon
return false;
}
int menu() //function for operation choice and execution
{
string choice; //CKE: change variable type here from int to string
do
{
cout << "\nPlease select an operation:\n" << endl
<< " 1 - Balance inquiry\n"
<< " 7 - Return card\n"
<< "\nEnter your choice and press return: ";
int balance = 512;
cin >> choice;
if (choice == "1" and isNumber(choice)) { cout << "Your balance is $" << balance << "\n\n"; } //CKE: semicolon replaced by output stream operator
else if (choice == "7" and isNumber(choice)) { cout << "Please wait...\nHave a good day." << endl; return 0; }
else { cout << "Invalid choice of operation."; } //CKE: remove recursion here as it isn't required
} while (!ReturnCard()); //CKE: negate result of ReturnCard function
cout << "Please wait...\nHave a good day." << endl;
return 0;
}
int main()
{
string choice;
cout << "Insert debit card to get started." << endl;
menu();
return 0;
}
I'm trying to create a small restaurant program in which I'll be practicing everything I learned in C++ so far. However I jumped into a small issue. At the beginning of the program, I prompt the user whether they want to enter the program, or leave it by choosing Y or N. If the input is anything other than that the program will tell the user is invalid.
The issue is lets say the user input one invalid character a.
The invalid output will be displayed normally and everything seems perfect.
But if the user inputs two characters, or more, the invalid output case will be printed as many as the characters input by the user. Sample below:
Output image
#include <iostream>
int main()
{
char ContinueAnswer;
std::string Employee {"Lara"};
std::cout << "\n\t\t\t---------------------------------------"
<< "\n\t\t\t| |"
<< "\n\t\t\t| Welcome to OP |"
<< "\n\t\t\t|Home to the best fast food in Orlando|"
<< "\n\t\t\t| |"
<< "\n\t\t\t--------------------------------------|" << std::endl;
do
{
std::cout << "\n\t\t\t Would you like to enter? (Y/N)"
<< "\n\t\t\t "; std::cin >> ContinueAnswer;
if(ContinueAnswer == 'y' || ContinueAnswer == 'Y')
{
system("cls");
std::cout << "\n\t\t\t My name is " << Employee << "."
<< "\n\t\t\tI will assist you as we go through the menu." << std::endl;
}
else if(ContinueAnswer == 'n' || ContinueAnswer == 'N')
{
std::cout << "\t\t\t\tGoodbye and come again!" << std::endl;
return 0;
}
else
std::cout << "\n\t\t\t\t Invalid Response" << std::endl;
}
while(ContinueAnswer != 'y' && ContinueAnswer != 'Y')
Thank you for taking time to read and for anyone who answers :)
You could simply make the user input a string:
std::string ContinueAnswer;
and compare like this:
if(ContinueAnswer == "y" || ContinueAnswer == "Y")
which will handle multi-character inputs.
If you want to handle spaces in the input as well, change the:
std::cin >> ContinueAnswer;
to:
std::getline(std::cin, ContinueAnswer);
Before addressing your question I need to point out that you should always verify that the input was successful before doing anything with it. Processing variables which were not set due to the inout failing is a rather common source of errors. For example:
if (std::cin >> ContinueAnswer) {
// do something with successfully read data
}
else {
// deal with the input failing, e.g., bail out
}
I assume you consider everything on the same line to be invalid if nine of the expected characters was read. You could read a line into an std::string. However, that could be abused to provide an extremely long line of input which would eventually crash your program. Also, reading data into a std::string just to throw it away seems ill-advised. I’d recommend ignoring all characters up to and including a newline which could be done using (you need to include <limits> for this approach):
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), ‘\n’);
The first argument is a special value indicating that there may be an arbitrary amount of character before the newline. In practice you could probably use a value like 1000 and it would be fine but it can be gamed. Of course, in a real application a dedicated limit may be used to prevent an adversary to keep the program busy for long. I tend to assume my programs are under attack to make sure I deal with unusual cases.
A quick refactor produces this:
#include <iostream>
#include <cstring>
#include <stdio.h>
int main()
{
char ContinueAnswer[256];
std::string Employee {"Lara"};
std::cout << "\n\t\t\t---------------------------------------"
<< "\n\t\t\t| |"
<< "\n\t\t\t| Welcome to OP |"
<< "\n\t\t\t|Home to the best fast food in Orlando|"
<< "\n\t\t\t| |"
<< "\n\t\t\t--------------------------------------|" << std::endl;
do
{
std::cout << "\n\t\t\t Would you like to enter? (Y/N)"
<< "\n\t\t\t "; std::cin.getline(ContinueAnswer,sizeof(ContinueAnswer));
if(strcmp(ContinueAnswer, "Y") == 0 || strcmp(ContinueAnswer, "y") == 0)
{
system("cls");
std::cout << "\n\t\t\t My name is " << Employee << "."
<< "\n\t\t\tI will assist you as we go through the menu." << std::endl;
}
else if(strcmp(ContinueAnswer, "N") == 0 || strcmp(ContinueAnswer, "n") == 0)
{
std::cout << "\t\t\t\tGoodbye and come again!" << std::endl;
return 0;
}
else
std::cout << "\n\t\t\t\t Invalid Response" << std::endl;
}
while(true);
}
The cin.getline will get all characters until a delimiter. Then, you can check for equivalence using strcmp and reject anything other than what you want. Lastly, it seems like you are wanting this to be in an infinite loop, so don't worry about checking the input at the end and just loop back.
I'm a beginner in programming, and I'm trying to make a program that calculated how much radiation you've been exposed to throughout your life. For some reason, the 'cin' in my xray function doesn't accept user input, and just exits with code 0.
#include <iostream>
#include <conio.h>
#include <windows.h>
#include <stdlib.h>
#include <string>
#include <sstream>
using namespace std;
bool nearpowerplant;
int XRay; // the amount of times you got an x-ray
double tRads = 0; // your total dose of radiation in your lifetime, measured in mSv (millisievert)
int age;
//the sleep function
void sleep() {
Sleep(1000); // 1000 miliseconds = 1 second
}
/*
>system("CLS")< for clear the console
*/
//introduction and pretty much the menu
void intro() {
cout << "Welcome to the Radiation Level Calculator" << endl;
sleep();
cout << "Conceptualized and created by Anatoly Zavyalov" << endl;
sleep();
cout << "Press the ENTER key to begin." << endl;
cin.get();
}
//introduction to general questions
void genintro() {
// intro to the medical
system("CLS");
sleep();
cout << "Let's begin with general questions." << endl;
sleep();
cout << "Press the ENTER key to continue." << endl;
cin.get();
}
//medical questions
void Age() {
//age
system("CLS");
cout << "How old are you?\n" << endl;
sleep();
cin >> age;
if (age <= 0) {
cout << "Your age can't be less or equal to 0." << endl;
Age();
}
else {
tRads += (age * 2);
sleep();
cout << tRads << endl;
}
}
//live close to powerplant?
void powerplant() {
system("CLS");
cout << "Do you live within 75 kilometers of a nuclear powerplant?" << endl;
sleep();
cout << "If yes, type YES. If no, type NO." << endl;
cin >> nearpowerplant;
if (nearpowerplant = "YES") {
tRads += (age * 0.01);
}
else {}
sleep();
cout << tRads << endl;
}
void xray() {
system("CLS");
cout << "How many times have you had an x-ray?\n" << endl;
sleep();
cin >> XRay;
if (XRay < 0) {
cout << "You can't have an x-ray a negative amount of times." << endl;
}
else {
tRads += (XRay * 3.1);
}
sleep();
cout << tRads << endl;
}
//main function, put all of the loops into here
int main() {
intro(); // the introduction
genintro(); // medical intro
Age(); // asks for age
powerplant(); // asks if lives close to powerplant
xray(); // asks for x-ray
return 0;
}
EDIT: I have edited the post to include the whole code. By the way, I am using Visual Studio Community 2017.
bool nearpowerplant;
nearpowerplant is a bool. It is true or false. That is it. It's worth noting that there is no reason for this variable to be globally accessible and consuming storage for the entire run of the program. It is used twice in the program, both times in the same function. It should be an Automatic variable scoped by the function that uses it.
cout << "If yes, type YES. If no, type NO." << endl;
cin >> nearpowerplant;
Reading "YES" or "NO" into a variable of type bool fails. cin cannot convert the string input into a boolean value and cin stops accepting input until the error is cleared. It's also a good idea to remove the garbage input that caused cin to fail or guess what? cin's just going to fail again. There are hundreds of SO questions on how to handle this, so I'm just going to drop keywords here: clear and ignore.
Takeaways: Make sure the data entry matches the type of the data being entered into and test the stream after every read to make sure the read succeeded.
eg:
if (cin >> nearpowerplant)
{
// do stuff
}
else
{
// clean up
}
This solves OP's visible error, but since it is heavily entwined with the next bug they are likely to find, we might as well cover it as well.
if (nearpowerplant = "YES") {
tRads += (age * 0.01);
}
else {}
if (nearpowerplant = "YES") { uses = (assignment) where it should use == (comparison). C++ is unforgiving here because this will compile. What it really did was takes the address of the string literal "YES", test that it's not null, and set nearpowerplant to the result. Since the address of the string literal is never going to be NULL, the result is always true, and when the if tests the result, the if will always enter.
Eg: http://ideone.com/4QL2jn
So what we need is something more like
cout << "If yes, type YES. If no, type NO." << endl;
string temp;
cin >> temp;
if (temp == "YES") {
tRads += (age * 0.01);
}
else {}
Note this will skip if the user inputs "yes", "y", "Yes" or anything other than exactly "YES". How you deal with this is up to you, but std::tolower and std::transform may help somewhat.
I think with sleep() comes undefined behavior, you should test it without, the os handles user-input and you do not have to care about the user typing in. endl flushes cout, so the text is directly shown.
Edit:
Maybe system("CLS") or sleep produces a silent error.
I'm new to C++. I decided to not watch the next tutorial and put my skills to use, by making a funny Mind Reader application. I'm pleased with myself, however, even though I've ironed out most bugs, I still have one concerning the exit function. I read the C++ documentation for it, and I'm not sure what I did wrong. I did exit(0);. I have a very weird error, which is:
no match for call to '(std::string {aka std::basic_string<char>}) (int)
I have searched online, however I am still unaware of what the problem is. My error is on line 59 (marked in the code):
#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
int main()
{
//declaring variables to be used later
string name;
string country;
int age;
//header goes below
cout << "#######################################";
" ############ MIND READER ############"
"#######################################\n\n";
//asks if the user would like to continue and in not, terminates
cout << "Would like you to have your mind read? Enter y for yes and n for no." << endl;
cout << "If you do not choose to proceed, this program will terminate." << endl;
string exitOrNot;
//receives user's input
cin >> exitOrNot;
//deals with input if it is 'y'
if (exitOrNot == "y"){
cout << "Okay, first you will need to sync your mind with this program. You will have to answer the following questions to synchronise.\n\n";
//asks questions
cout << "Firstly, please enter your full name, with correct capitalisation:\n\n";
cin >> name;
cout << "Now please enter the country you are in at the moment:\n\n";
cin >> country;
cout << "This will be the final question; please provide your age:\n\n";
cin >> age;
//asks the user to start the sync
cout << "There is enough information to start synchronisation. Enter p to start the sync...\n\n";
string proceed;
cin >> proceed;
//checks to see if to proceed and does so
if (proceed == "p"){
//provides results of mind read
cout << "Sync complete." << endl;
cout << "Your mind has been synced and read.\n\n";
cout << "However, due to too much interference, only limited data was aquired from your mind." << endl;
cout << "Here is what was read from your mind:\n\n";
//puts variables in sentence
cout << "Your name is " << name << " and you are " << age << " years old. You are based in " << country << "." << endl << "\n\n";
cout << "Thanks for using Mind Reader, have a nice day. Enter e to exit." << endl;
//terminates the program the program
string exit;
cin >> exit;
if (exit == "e"){
exit(0); // <------------- LINE 59
}
}
}
//terminates the program if the input is 'n'
if (exitOrNot == "n"){
exit(0);
}
return 0;
}
Thanks
The local variable exit shadows other identifiers from outer scopes with the same name.
To illustrate with a smaller example:
int main()
{
int i;
{
int i;
i = 0; // assign to the "nearest" i
// the other i cannot be reached from this scope
}
}
Since the only exit visible is an object of type std::string, the compiler sees exit(0) as a call to operator()(int) and throws a hissy fit when it doesn't find one among std::string members.
You can either qualify the name (std::exit(0);) or rename the variable. And since all of your code is in main you can simply say return 0; instead.
Try using return 0; or return EXIT_SUCCESS;. It's the exact same thing. Also, you can only input one word into a cin. Instead, use getline(cin, string name); If it still doesn't work, add a cin.ignore(); before your getline(cin, string name);, like this:
//stuff
string country;
cout << "Now please enter the country you are in at the moment:\n\n";
cin.ignore();
getline(cin, country);
//stuff
return 0;
The problem is arrising because you declared a standard keyword as the name of a local variable.
Now as the local variable is of type sting it is not able to take it as its value.
I have a while statement that keeps repeating the text without giving the user a chance to input another value for action. What am I doing wrong? It still doesn't ask for input. I need for the code to display the text once, then ask for input. Presumably, if you typed anything but 1 it would repeat the sequence. But as it stands it simply kicks you out of the loop without the chance to correct the action (As of the last edit, see below.)
int action = 0;
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
cin >> action;
}
One suggestion was:
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
cin >> action;
cin.ignore();
}
That still produces text over and over.
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
if (!(cin >> action))
// ...problems in the I/O stream...
break;
}
This one kicks you out without a chance to input a new action.
If you type a character that is not white space and can't be part of an integer, then you have an infinite loop. Each attempt to input to action fails on the invalid character without changing the value stored in action.
You could write:
int action = 0;
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
if (!(cin >> action))
// ...problems in the I/O stream...
break;
}
This will handle EOF and alphabetic characters more gracefully than a continuous loop. You might need to set a flag or return an error condition from the function or do something else other than break out of the loop. Always check your inputs for success.
You might also consider outputting the value you're getting stored in action in the loop, so you can see what is happening:
int action = 0;
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
if (!(cin >> action))
// ...problems in the I/O stream...
break;
cerr << "Action: " << action << endl;
}
This might tell you something useful too.
Please show a complete little program that illustrates your problem — an SSCCE (Short, Self-Contained, Correct Example).
For example, I'm testing with:
#include <iostream>
using namespace std;
int main()
{
int action = 0;
while (action != 1)
{
cout << " No you must look it might be dangerous" << endl;
if (!(cin >> action))
{
// ...problems in the I/O stream...
break;
}
cout << "Action: " << action << endl;
}
cout << "After loop" << endl;
if (!cin)
cout << "cin is bust" << endl;
else
cout << "Action: " << action << endl;
}
That's no longer minimal code — the material after the loop is merely telling me what is happening. But it does help me ensure that my code is doing what I expect.
What does your equivalent code look like, and what are you typing in response to the prompts — and especially, what are you typing before you get to this code fragment (and what other input activity is going on before you get here)?