How to update value of variable declared outside lambda function?

Question

Came across the following code today, it converts the input seconds in Long to a format like 2days, 3hours, 1min, 5s. My problem with this is the use of final long[] secondsCpy = { seconds };. Any variable inside a lambda has to be final or effectively final, so, using an array variable is kind of a hack. Is there a better way to do this?

    private static final LinkedHashMap<String, Long> readableTimeFormatMap = new LinkedHashMap<String, Long>() {
        {
           put("day", TimeUnit.DAYS.toSeconds(1));
           put("hr", TimeUnit.HOURS.toSeconds(1));
           put("min", TimeUnit.MINUTES.toSeconds(1));
           put("sec", TimeUnit.SECONDS.toSeconds(1));
        }
     };


    public static String getReadableTime(final long seconds) {

        final StringJoiner readableTime = new StringJoiner(" ");

        final long[] secondsCpy = { seconds };

        readableTimeFormatMap.forEach((displayString, divider) -> {
            readableTime.add(getReadableTimeUnit(displayString, secondsCpy[0] / divider));
            secondsCpy[0] = secondsCpy[0] % divider;
        });

        return readableTime.toString().trim();
    }

return any value from your lambda with a combination of map and collect, instead of using foreach --> no need of any final variable this way! — HBo
– HBo, Commented Apr 18, 2018 at 14:44

Kermi · Accepted Answer · 2018-04-18 14:45:56Z

3

There is no much better way to do this, you can use for example AtomicLong

answered Apr 18, 2018 at 14:45

Kermi

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Comments

Ousmane D. · Accepted Answer · 2018-04-18 15:54:47Z

2

Well without complicating things, you could always resort to using an enhanced for loop.

long secondsCpy = seconds;
for (Map.Entry<String, Long> entry : map.entrySet()) {
    readableTime.add(getReadableTimeUnit(entry.getKey(), secondsCpy / entry.getValue()));
    secondsCpy = secondsCpy % entry.getValue();
}

answered Apr 18, 2018 at 15:54

Ousmane D.

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4 Comments

Holger Over a year ago

You don’t even need the secondsCpy, as seconds will not be used for anything else in this method.

Kermi Over a year ago

@Holger in the question you can see that seconds is a final variable, so it is necessary to do a copy.

Harish Over a year ago

Yeah, we could use this too, but wanted to see if it can be done using a lambda function. Thanks!

Holger Over a year ago

@Kermi or you remove that final modifier. It has no practical relevance anyway (besides making a second variable necessary).

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