java: Trying to return a string value from an array of strings via a method

Question

public String GetValue(String key) {
    for (int i = 1; i<=length; i++) {
        if (key.equals(keyArr[i])) {
            return valueArr[i];
        }
    }
}
public String GetKey(String value) {
    String key;
    for (int i = 1; i<=length; i++) {
        if (value.equals(valueArr[i])) {
            key = keyArr[i];
            return key;
        }
    }
}

I get an error "this method must return a result of type string". But the two arrays valueArr and keyArr are both String type arrays. I know that the value of keyArr[i] and valueArr[i] are strings because if I change the return type of the method to something else it says that it expected something else and got a string.

Does this even compile? It should be complaining about a missing return statement. — Alex S
– Alex S, Commented Jan 29, 2017 at 1:52

Darshan Mehta · Accepted Answer · 2017-01-29 00:04:05Z

1

That's because it does not always return a String. E.g. if the key is not found, if (key.equals(keyArr[i])) { will never be true and hence, return valueArr[i]; will never get executed.

To fix this, you need to do the following:

Add return statement after for loop, to return something if no match is found, e.g.:

public String GetValue(String key) {
    for (int i = 0; i < length; i++) {
        if (key.equals(keyArr[i])) {
            return valueArr[i];
        }
    }
    return null;
}

Handle the return type in the calling method (i.e. add a null check or perform an appropriate action if no match found).

edited Jan 29, 2017 at 0:04

answered Jan 28, 2017 at 23:58

Darshan Mehta

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6 Comments

Mohsen_Fatemi Over a year ago

you can't start indexing from 1 to <=length

Darshan Mehta Over a year ago

@Mohsen_Fatemi corrected, thanks for pointing it out. I actually used the example from question only.

Mohsen_Fatemi Over a year ago

i don't know how does it work without ArrayIndexOutOfBoundException for @Johndoe ? :))

Mohsen_Fatemi Over a year ago

@DarshanMehta yw ;)

John doe Over a year ago

i meant for that to be a zero, i<=length-1 would work as well thanks

|

templatetypedef · Accepted Answer · 2017-01-28 23:57:14Z

1

The issue less with the "of type String" and more with the "must." What happens, for example, if you pass in a string that doesn't appear in the array? In that case, your function never actually returns anything.

If you're guaranteed that the key will always exist, consider putting a line at the end like

throw new RuntimeException("Key not found?");

so that the function doesn't just drop off the end. Alternatively, try adding a line like

return null;

so that you explicitly return something in that case.

answered Jan 28, 2017 at 23:57

templatetypedef

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4 Comments

Grzegorz Górkiewicz Over a year ago

return null may not be the best option. Why not something like -1? It is already used in some methods like binarySearch on an array.

Aloso Over a year ago

Why not return -1? Because return type is String. Apart from that, @templatetypedef also offered a different, perhaps preferable solution using an Exception.

Grzegorz Górkiewicz Over a year ago

Exception is better here... But returning null can become really problematic.

templatetypedef Over a year ago

@GrzegorzGórkiewicz I agree that null might not be the best return value here. That was mostly showing off another way to get the error to stop, which was by always returning a value.

Mohsen_Fatemi · Accepted Answer · 2017-01-28 23:59:15Z

0

the problem is when you Write the type String you must return a String value in your method. so for Example GetValue Could be like this :

public String GetValue(String key) {
    for (int i = 0; i<keyArr.length; i++) {
        if (key.equals(keyArr[i])) {
            return valueArr[i];
        }
    }
    return "";
}

pay attention to indexes , it must start from 0 to length -1 in arrays

and your second method :

public String GetKey(String value) {
    String key = "";
    for (int i = 0; i<value.length; i++) {
        if (value.equals(valueArr[i])) {
            key = keyArr[i];
            return key;
        }
    }
    return key ;
}

answered Jan 28, 2017 at 23:59

Mohsen_Fatemi

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