Java, “Variable name” cannot be resolved to a variable











up vote
15
down vote

favorite
3












I use Eclipse using Java, I get this error:



"Variable name" cannot be resolved to a variable.


With this Java program:



public class SalCal {
private int hoursWorked;
public SalCal(String name, int hours, double hoursRate) {
nameEmployee = name;
hoursWorked = hours;
ratePrHour = hoursRate;
}
public void setHoursWorked() {
hoursWorked = hours; //ERROR HERE, hours cannot be resolved to a type
}
public double calculateSalary() {
if (hoursWorked <= 40) {
totalSalary = ratePrHour * (double) hoursWorked;
}
if (hoursWorked > 40) {
salaryAfter40 = hoursWorked - 40;
totalSalary = (ratePrHour * 40)
+ (ratePrHour * 1.5 * salaryAfter40);
}
return totalSalary;
}
}


What causes this error message?










share|improve this question




























    up vote
    15
    down vote

    favorite
    3












    I use Eclipse using Java, I get this error:



    "Variable name" cannot be resolved to a variable.


    With this Java program:



    public class SalCal {
    private int hoursWorked;
    public SalCal(String name, int hours, double hoursRate) {
    nameEmployee = name;
    hoursWorked = hours;
    ratePrHour = hoursRate;
    }
    public void setHoursWorked() {
    hoursWorked = hours; //ERROR HERE, hours cannot be resolved to a type
    }
    public double calculateSalary() {
    if (hoursWorked <= 40) {
    totalSalary = ratePrHour * (double) hoursWorked;
    }
    if (hoursWorked > 40) {
    salaryAfter40 = hoursWorked - 40;
    totalSalary = (ratePrHour * 40)
    + (ratePrHour * 1.5 * salaryAfter40);
    }
    return totalSalary;
    }
    }


    What causes this error message?










    share|improve this question


























      up vote
      15
      down vote

      favorite
      3









      up vote
      15
      down vote

      favorite
      3






      3





      I use Eclipse using Java, I get this error:



      "Variable name" cannot be resolved to a variable.


      With this Java program:



      public class SalCal {
      private int hoursWorked;
      public SalCal(String name, int hours, double hoursRate) {
      nameEmployee = name;
      hoursWorked = hours;
      ratePrHour = hoursRate;
      }
      public void setHoursWorked() {
      hoursWorked = hours; //ERROR HERE, hours cannot be resolved to a type
      }
      public double calculateSalary() {
      if (hoursWorked <= 40) {
      totalSalary = ratePrHour * (double) hoursWorked;
      }
      if (hoursWorked > 40) {
      salaryAfter40 = hoursWorked - 40;
      totalSalary = (ratePrHour * 40)
      + (ratePrHour * 1.5 * salaryAfter40);
      }
      return totalSalary;
      }
      }


      What causes this error message?










      share|improve this question















      I use Eclipse using Java, I get this error:



      "Variable name" cannot be resolved to a variable.


      With this Java program:



      public class SalCal {
      private int hoursWorked;
      public SalCal(String name, int hours, double hoursRate) {
      nameEmployee = name;
      hoursWorked = hours;
      ratePrHour = hoursRate;
      }
      public void setHoursWorked() {
      hoursWorked = hours; //ERROR HERE, hours cannot be resolved to a type
      }
      public double calculateSalary() {
      if (hoursWorked <= 40) {
      totalSalary = ratePrHour * (double) hoursWorked;
      }
      if (hoursWorked > 40) {
      salaryAfter40 = hoursWorked - 40;
      totalSalary = (ratePrHour * 40)
      + (ratePrHour * 1.5 * salaryAfter40);
      }
      return totalSalary;
      }
      }


      What causes this error message?







      java variables name-resolution






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Sep 15 '13 at 3:47









      Eric Leschinski

      84.6k36316270




      84.6k36316270










      asked Sep 28 '11 at 19:50









      user820913

      3282718




      3282718
























          3 Answers
          3






          active

          oldest

          votes

















          up vote
          10
          down vote



          accepted










          If you look at the scope of the variable 'hoursWorked' you will see that it is a member of the class (declared as private int)



          The two variables you are having trouble with are passed as parameters to the constructor.



          The error message is because 'hours' is out of scope in the setter.






          share|improve this answer






























            up vote
            8
            down vote













            public void setHoursWorked(){
            hoursWorked = hours;
            }


            You haven't defined hours inside that method. hours is not passed in as a parameter, it's not declared as a variable, and it's not being used as a class member, so you get that error.






            share|improve this answer






























              up vote
              3
              down vote













              I've noticed bizarre behavior with Eclipse version 4.2.1 delivering me this error:



              String cannot be resolved to a variable


              With this Java code:



              if (true)
              String my_variable = "somevalue";
              System.out.println("foobar");


              You would think this code is very straight forward, the conditional is true, we set my_variable to somevalue. And it should print foobar. Right?



              Wrong, you get the above mentioned compile time error. Eclipse is trying to prevent you from making a mistake by assuming that both statements are within the if statement.



              If you put braces around the conditional block like this:



              if (true){
              String my_variable = "somevalue"; }
              System.out.println("foobar");


              Then it compiles and runs fine. Apparently poorly bracketed conditionals are fair game for generating compile time errors now.






              share|improve this answer




















                protected by Community May 22 at 6:51



                Thank you for your interest in this question.
                Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



                Would you like to answer one of these unanswered questions instead?














                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                10
                down vote



                accepted










                If you look at the scope of the variable 'hoursWorked' you will see that it is a member of the class (declared as private int)



                The two variables you are having trouble with are passed as parameters to the constructor.



                The error message is because 'hours' is out of scope in the setter.






                share|improve this answer



























                  up vote
                  10
                  down vote



                  accepted










                  If you look at the scope of the variable 'hoursWorked' you will see that it is a member of the class (declared as private int)



                  The two variables you are having trouble with are passed as parameters to the constructor.



                  The error message is because 'hours' is out of scope in the setter.






                  share|improve this answer

























                    up vote
                    10
                    down vote



                    accepted







                    up vote
                    10
                    down vote



                    accepted






                    If you look at the scope of the variable 'hoursWorked' you will see that it is a member of the class (declared as private int)



                    The two variables you are having trouble with are passed as parameters to the constructor.



                    The error message is because 'hours' is out of scope in the setter.






                    share|improve this answer














                    If you look at the scope of the variable 'hoursWorked' you will see that it is a member of the class (declared as private int)



                    The two variables you are having trouble with are passed as parameters to the constructor.



                    The error message is because 'hours' is out of scope in the setter.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Sep 15 '13 at 3:49









                    Eric Leschinski

                    84.6k36316270




                    84.6k36316270










                    answered Sep 28 '11 at 19:59









                    Hugh Jones

                    2,133926




                    2,133926
























                        up vote
                        8
                        down vote













                        public void setHoursWorked(){
                        hoursWorked = hours;
                        }


                        You haven't defined hours inside that method. hours is not passed in as a parameter, it's not declared as a variable, and it's not being used as a class member, so you get that error.






                        share|improve this answer



























                          up vote
                          8
                          down vote













                          public void setHoursWorked(){
                          hoursWorked = hours;
                          }


                          You haven't defined hours inside that method. hours is not passed in as a parameter, it's not declared as a variable, and it's not being used as a class member, so you get that error.






                          share|improve this answer

























                            up vote
                            8
                            down vote










                            up vote
                            8
                            down vote









                            public void setHoursWorked(){
                            hoursWorked = hours;
                            }


                            You haven't defined hours inside that method. hours is not passed in as a parameter, it's not declared as a variable, and it's not being used as a class member, so you get that error.






                            share|improve this answer














                            public void setHoursWorked(){
                            hoursWorked = hours;
                            }


                            You haven't defined hours inside that method. hours is not passed in as a parameter, it's not declared as a variable, and it's not being used as a class member, so you get that error.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Sep 15 '13 at 3:49









                            Eric Leschinski

                            84.6k36316270




                            84.6k36316270










                            answered Sep 28 '11 at 19:55









                            Marc B

                            311k31317416




                            311k31317416






















                                up vote
                                3
                                down vote













                                I've noticed bizarre behavior with Eclipse version 4.2.1 delivering me this error:



                                String cannot be resolved to a variable


                                With this Java code:



                                if (true)
                                String my_variable = "somevalue";
                                System.out.println("foobar");


                                You would think this code is very straight forward, the conditional is true, we set my_variable to somevalue. And it should print foobar. Right?



                                Wrong, you get the above mentioned compile time error. Eclipse is trying to prevent you from making a mistake by assuming that both statements are within the if statement.



                                If you put braces around the conditional block like this:



                                if (true){
                                String my_variable = "somevalue"; }
                                System.out.println("foobar");


                                Then it compiles and runs fine. Apparently poorly bracketed conditionals are fair game for generating compile time errors now.






                                share|improve this answer

























                                  up vote
                                  3
                                  down vote













                                  I've noticed bizarre behavior with Eclipse version 4.2.1 delivering me this error:



                                  String cannot be resolved to a variable


                                  With this Java code:



                                  if (true)
                                  String my_variable = "somevalue";
                                  System.out.println("foobar");


                                  You would think this code is very straight forward, the conditional is true, we set my_variable to somevalue. And it should print foobar. Right?



                                  Wrong, you get the above mentioned compile time error. Eclipse is trying to prevent you from making a mistake by assuming that both statements are within the if statement.



                                  If you put braces around the conditional block like this:



                                  if (true){
                                  String my_variable = "somevalue"; }
                                  System.out.println("foobar");


                                  Then it compiles and runs fine. Apparently poorly bracketed conditionals are fair game for generating compile time errors now.






                                  share|improve this answer























                                    up vote
                                    3
                                    down vote










                                    up vote
                                    3
                                    down vote









                                    I've noticed bizarre behavior with Eclipse version 4.2.1 delivering me this error:



                                    String cannot be resolved to a variable


                                    With this Java code:



                                    if (true)
                                    String my_variable = "somevalue";
                                    System.out.println("foobar");


                                    You would think this code is very straight forward, the conditional is true, we set my_variable to somevalue. And it should print foobar. Right?



                                    Wrong, you get the above mentioned compile time error. Eclipse is trying to prevent you from making a mistake by assuming that both statements are within the if statement.



                                    If you put braces around the conditional block like this:



                                    if (true){
                                    String my_variable = "somevalue"; }
                                    System.out.println("foobar");


                                    Then it compiles and runs fine. Apparently poorly bracketed conditionals are fair game for generating compile time errors now.






                                    share|improve this answer












                                    I've noticed bizarre behavior with Eclipse version 4.2.1 delivering me this error:



                                    String cannot be resolved to a variable


                                    With this Java code:



                                    if (true)
                                    String my_variable = "somevalue";
                                    System.out.println("foobar");


                                    You would think this code is very straight forward, the conditional is true, we set my_variable to somevalue. And it should print foobar. Right?



                                    Wrong, you get the above mentioned compile time error. Eclipse is trying to prevent you from making a mistake by assuming that both statements are within the if statement.



                                    If you put braces around the conditional block like this:



                                    if (true){
                                    String my_variable = "somevalue"; }
                                    System.out.println("foobar");


                                    Then it compiles and runs fine. Apparently poorly bracketed conditionals are fair game for generating compile time errors now.







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Sep 15 '13 at 3:59









                                    Eric Leschinski

                                    84.6k36316270




                                    84.6k36316270

















                                        protected by Community May 22 at 6:51



                                        Thank you for your interest in this question.
                                        Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



                                        Would you like to answer one of these unanswered questions instead?



                                        Popular posts from this blog

                                        Bressuire

                                        Vorschmack

                                        Quarantine