Java, "Variable name" cannot be resolved to a variable

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?

score 16 · Accepted Answer · edited Sep 15 '13 at 03:49

16

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.

edited Sep 15 '13 at 03:49

Eric Leschinski

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answered Sep 28 '11 at 19:59

Hugh Jones

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score 10 · Answer 2 · edited Sep 15 '13 at 03:49

10

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.

edited Sep 15 '13 at 03:49

Eric Leschinski

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answered Sep 28 '11 at 19:55

Marc B

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score 3 · Answer 3 · answered Sep 15 '13 at 03:59

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.

Java, "Variable name" cannot be resolved to a variable

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