Unable to check for non-integer input










1














The else statement in this code does not execute, instead, when a character is entered, it would get stuck in a continuous loop, repeating the lines 33 and 35.



I want to check if the user input is not an integer in the else statement and ask the user again to specify the age.



#include <iostream>
#include <windows.h>
using namespace std;

class User
string name;
public:
string getName()
return name;

void setName(string newName)
name = newName;

;

int main()

/** Initialise Variables */
string name;
int age;
User u1;
bool running = true;


while(running)
/** Welcome Message w/ User's Name */
cout << "What is your name?: " << endl;
cin >> name;
u1.setName(name);
cout << "Welcome, " << u1.getName() << "!" << endl;
returnToAge:
Sleep(1000);
cout << u1.getName() << ", to continue, you have to be over 18." << endl;
Sleep(1000);
cout << "Please enter your age: " << endl;
cin >> age;

if (age >= 18)
cout << "You may continue..." << endl;
Sleep(1000);
//Enter rest of questionnaire here



else if(age < 18)
cout << "You are underage! Please try again!" << endl;
Sleep(1500);
goto returnToAge;

else
cout << "INVALID INPUT!" << endl;
goto returnToAge;













share|improve this question





















  • You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
    – Rags
    Nov 12 at 11:29











  • A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
    – Henning Koehler
    Nov 12 at 11:30






  • 1




    @Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
    – Leon185
    Nov 12 at 11:32






  • 2




    A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
    – Christian G
    Nov 12 at 11:40






  • 1




    @ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
    – Rags
    Nov 12 at 11:50















1














The else statement in this code does not execute, instead, when a character is entered, it would get stuck in a continuous loop, repeating the lines 33 and 35.



I want to check if the user input is not an integer in the else statement and ask the user again to specify the age.



#include <iostream>
#include <windows.h>
using namespace std;

class User
string name;
public:
string getName()
return name;

void setName(string newName)
name = newName;

;

int main()

/** Initialise Variables */
string name;
int age;
User u1;
bool running = true;


while(running)
/** Welcome Message w/ User's Name */
cout << "What is your name?: " << endl;
cin >> name;
u1.setName(name);
cout << "Welcome, " << u1.getName() << "!" << endl;
returnToAge:
Sleep(1000);
cout << u1.getName() << ", to continue, you have to be over 18." << endl;
Sleep(1000);
cout << "Please enter your age: " << endl;
cin >> age;

if (age >= 18)
cout << "You may continue..." << endl;
Sleep(1000);
//Enter rest of questionnaire here



else if(age < 18)
cout << "You are underage! Please try again!" << endl;
Sleep(1500);
goto returnToAge;

else
cout << "INVALID INPUT!" << endl;
goto returnToAge;













share|improve this question





















  • You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
    – Rags
    Nov 12 at 11:29











  • A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
    – Henning Koehler
    Nov 12 at 11:30






  • 1




    @Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
    – Leon185
    Nov 12 at 11:32






  • 2




    A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
    – Christian G
    Nov 12 at 11:40






  • 1




    @ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
    – Rags
    Nov 12 at 11:50













1












1








1







The else statement in this code does not execute, instead, when a character is entered, it would get stuck in a continuous loop, repeating the lines 33 and 35.



I want to check if the user input is not an integer in the else statement and ask the user again to specify the age.



#include <iostream>
#include <windows.h>
using namespace std;

class User
string name;
public:
string getName()
return name;

void setName(string newName)
name = newName;

;

int main()

/** Initialise Variables */
string name;
int age;
User u1;
bool running = true;


while(running)
/** Welcome Message w/ User's Name */
cout << "What is your name?: " << endl;
cin >> name;
u1.setName(name);
cout << "Welcome, " << u1.getName() << "!" << endl;
returnToAge:
Sleep(1000);
cout << u1.getName() << ", to continue, you have to be over 18." << endl;
Sleep(1000);
cout << "Please enter your age: " << endl;
cin >> age;

if (age >= 18)
cout << "You may continue..." << endl;
Sleep(1000);
//Enter rest of questionnaire here



else if(age < 18)
cout << "You are underage! Please try again!" << endl;
Sleep(1500);
goto returnToAge;

else
cout << "INVALID INPUT!" << endl;
goto returnToAge;













share|improve this question













The else statement in this code does not execute, instead, when a character is entered, it would get stuck in a continuous loop, repeating the lines 33 and 35.



I want to check if the user input is not an integer in the else statement and ask the user again to specify the age.



#include <iostream>
#include <windows.h>
using namespace std;

class User
string name;
public:
string getName()
return name;

void setName(string newName)
name = newName;

;

int main()

/** Initialise Variables */
string name;
int age;
User u1;
bool running = true;


while(running)
/** Welcome Message w/ User's Name */
cout << "What is your name?: " << endl;
cin >> name;
u1.setName(name);
cout << "Welcome, " << u1.getName() << "!" << endl;
returnToAge:
Sleep(1000);
cout << u1.getName() << ", to continue, you have to be over 18." << endl;
Sleep(1000);
cout << "Please enter your age: " << endl;
cin >> age;

if (age >= 18)
cout << "You may continue..." << endl;
Sleep(1000);
//Enter rest of questionnaire here



else if(age < 18)
cout << "You are underage! Please try again!" << endl;
Sleep(1500);
goto returnToAge;

else
cout << "INVALID INPUT!" << endl;
goto returnToAge;










c++






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 12 at 11:25









Leon185

5711




5711











  • You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
    – Rags
    Nov 12 at 11:29











  • A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
    – Henning Koehler
    Nov 12 at 11:30






  • 1




    @Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
    – Leon185
    Nov 12 at 11:32






  • 2




    A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
    – Christian G
    Nov 12 at 11:40






  • 1




    @ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
    – Rags
    Nov 12 at 11:50
















  • You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
    – Rags
    Nov 12 at 11:29











  • A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
    – Henning Koehler
    Nov 12 at 11:30






  • 1




    @Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
    – Leon185
    Nov 12 at 11:32






  • 2




    A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
    – Christian G
    Nov 12 at 11:40






  • 1




    @ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
    – Rags
    Nov 12 at 11:50















You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
– Rags
Nov 12 at 11:29





You will probably find that not many people like the use of the goto. There tend to be better ways of structuring code so that they're not necessary (they're messy, and hard to control, especially as code grows.) Can I suggest you use a loop around the age entry, and only exit the loop when the conditions are met? Also, might be good to put that sort of operation into a function, as it can be separate from the main flow then, and easier to see/understand.
– Rags
Nov 12 at 11:29













A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
– Henning Koehler
Nov 12 at 11:30




A very general comment: don't use labels and goto statements. They are never necessary, and create hard to read code (and bugs as a result of that).
– Henning Koehler
Nov 12 at 11:30




1




1




@Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
– Leon185
Nov 12 at 11:32




@Rags I know there are better ways of using something other than goto. It is currently the most understandable code to me, to use at this moment in time. I will use something better as the code grows. I'll have a go with what you have said.
– Leon185
Nov 12 at 11:32




2




2




A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
– Christian G
Nov 12 at 11:40




A classic in Computer Science history: "goto statements considered harmful" by Edgar Dijkstra homepages.cwi.nl/~storm/teaching/reader/Dijkstra68.pdf
– Christian G
Nov 12 at 11:40




1




1




@ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
– Rags
Nov 12 at 11:50




@ChristianG Not only that - there's a lot by Dijkstra that's worth reading. Some of it is quite unpalatable to some people - but his insight was amazing.
– Rags
Nov 12 at 11:50












3 Answers
3






active

oldest

votes


















2














The problem is that cin >> age will read an integer into age (as that's the only thing age can store), and no matter what integer that is (it's 0 for non-numbers), it will be either bigger or smaller than 18, hence why your else case is never reached.



To detect invalid input, you need to read in a string and then check whether that string represents a valid integer.






share|improve this answer




















  • Ah! I understand now. Thanks :)
    – Leon185
    Nov 12 at 11:34


















2














You need to check if the input operation succeeded. The easiest way is:



if (cin >> age) 
// now use age
else
// error



Another way which is equivalent:



cin >> age;
if (!cin)
// error






share|improve this answer




























    1














    Use std::cin::fail to check wether the user input is an integer.



    Secondly, avoid maximum practice using gotos, as it produces Spaghetti code structure. You can replace the goto using a while loop as follows:



    while (true) 

    // code
    std::cin >> age;
    if (!std::cin.fail() && age >= 18) // if cin not failed && the codition

    /* do something */
    break; // break the loop!

    else if (!std::cin.fail() && age < 18) // if cin not failed && the codition

    /* do something goto removeed */

    else std::cout << "INVALID INPUT!" << std::endl; /* goto removeed */






    share|improve this answer




















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      2














      The problem is that cin >> age will read an integer into age (as that's the only thing age can store), and no matter what integer that is (it's 0 for non-numbers), it will be either bigger or smaller than 18, hence why your else case is never reached.



      To detect invalid input, you need to read in a string and then check whether that string represents a valid integer.






      share|improve this answer




















      • Ah! I understand now. Thanks :)
        – Leon185
        Nov 12 at 11:34















      2














      The problem is that cin >> age will read an integer into age (as that's the only thing age can store), and no matter what integer that is (it's 0 for non-numbers), it will be either bigger or smaller than 18, hence why your else case is never reached.



      To detect invalid input, you need to read in a string and then check whether that string represents a valid integer.






      share|improve this answer




















      • Ah! I understand now. Thanks :)
        – Leon185
        Nov 12 at 11:34













      2












      2








      2






      The problem is that cin >> age will read an integer into age (as that's the only thing age can store), and no matter what integer that is (it's 0 for non-numbers), it will be either bigger or smaller than 18, hence why your else case is never reached.



      To detect invalid input, you need to read in a string and then check whether that string represents a valid integer.






      share|improve this answer












      The problem is that cin >> age will read an integer into age (as that's the only thing age can store), and no matter what integer that is (it's 0 for non-numbers), it will be either bigger or smaller than 18, hence why your else case is never reached.



      To detect invalid input, you need to read in a string and then check whether that string represents a valid integer.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Nov 12 at 11:33









      Henning Koehler

      1,129610




      1,129610











      • Ah! I understand now. Thanks :)
        – Leon185
        Nov 12 at 11:34
















      • Ah! I understand now. Thanks :)
        – Leon185
        Nov 12 at 11:34















      Ah! I understand now. Thanks :)
      – Leon185
      Nov 12 at 11:34




      Ah! I understand now. Thanks :)
      – Leon185
      Nov 12 at 11:34













      2














      You need to check if the input operation succeeded. The easiest way is:



      if (cin >> age) 
      // now use age
      else
      // error



      Another way which is equivalent:



      cin >> age;
      if (!cin)
      // error






      share|improve this answer

























        2














        You need to check if the input operation succeeded. The easiest way is:



        if (cin >> age) 
        // now use age
        else
        // error



        Another way which is equivalent:



        cin >> age;
        if (!cin)
        // error






        share|improve this answer























          2












          2








          2






          You need to check if the input operation succeeded. The easiest way is:



          if (cin >> age) 
          // now use age
          else
          // error



          Another way which is equivalent:



          cin >> age;
          if (!cin)
          // error






          share|improve this answer












          You need to check if the input operation succeeded. The easiest way is:



          if (cin >> age) 
          // now use age
          else
          // error



          Another way which is equivalent:



          cin >> age;
          if (!cin)
          // error







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 12 at 11:35









          John Zwinck

          150k16175286




          150k16175286





















              1














              Use std::cin::fail to check wether the user input is an integer.



              Secondly, avoid maximum practice using gotos, as it produces Spaghetti code structure. You can replace the goto using a while loop as follows:



              while (true) 

              // code
              std::cin >> age;
              if (!std::cin.fail() && age >= 18) // if cin not failed && the codition

              /* do something */
              break; // break the loop!

              else if (!std::cin.fail() && age < 18) // if cin not failed && the codition

              /* do something goto removeed */

              else std::cout << "INVALID INPUT!" << std::endl; /* goto removeed */






              share|improve this answer

























                1














                Use std::cin::fail to check wether the user input is an integer.



                Secondly, avoid maximum practice using gotos, as it produces Spaghetti code structure. You can replace the goto using a while loop as follows:



                while (true) 

                // code
                std::cin >> age;
                if (!std::cin.fail() && age >= 18) // if cin not failed && the codition

                /* do something */
                break; // break the loop!

                else if (!std::cin.fail() && age < 18) // if cin not failed && the codition

                /* do something goto removeed */

                else std::cout << "INVALID INPUT!" << std::endl; /* goto removeed */






                share|improve this answer























                  1












                  1








                  1






                  Use std::cin::fail to check wether the user input is an integer.



                  Secondly, avoid maximum practice using gotos, as it produces Spaghetti code structure. You can replace the goto using a while loop as follows:



                  while (true) 

                  // code
                  std::cin >> age;
                  if (!std::cin.fail() && age >= 18) // if cin not failed && the codition

                  /* do something */
                  break; // break the loop!

                  else if (!std::cin.fail() && age < 18) // if cin not failed && the codition

                  /* do something goto removeed */

                  else std::cout << "INVALID INPUT!" << std::endl; /* goto removeed */






                  share|improve this answer












                  Use std::cin::fail to check wether the user input is an integer.



                  Secondly, avoid maximum practice using gotos, as it produces Spaghetti code structure. You can replace the goto using a while loop as follows:



                  while (true) 

                  // code
                  std::cin >> age;
                  if (!std::cin.fail() && age >= 18) // if cin not failed && the codition

                  /* do something */
                  break; // break the loop!

                  else if (!std::cin.fail() && age < 18) // if cin not failed && the codition

                  /* do something goto removeed */

                  else std::cout << "INVALID INPUT!" << std::endl; /* goto removeed */







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 12 at 11:46









                  JeJo

                  3,9493625




                  3,9493625



























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