Java Map computeIfAbsent issue

So I found a curiosity with Java's Map computeIfAbsent (using java8) method and I hope someone can tell me why this is happening as I can not really follow the logic behind that issue.

So, I have a Map with a key (obviously) and the value is a list and I use the computeIfAbsent to create a new list when a key is not set yet. Now when I use an Integer as key I can use the following:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

But when I use a String as key trying to use

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

I get the error that The method computeIfAbsent(String, Function<? super String,? extends List<Object>>) in the type Map<String,List<Object>> is not applicable for the arguments (String, ArrayList::new). Is there just the implementation missing? Using a String key I have to use the method like that, which is then working again.

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>());

Maybe someone can enlighten me about that. Thanks :)

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

As a side note, Guava's MultiMap already implements what you're looking for.
– chrylis
Dec 19 '18 at 7:28

and its worth mentioning the type of map used in the code for clarity of the question
– nullpointer
Dec 19 '18 at 7:46

I'm pretty sure this is a duplicate, but it's hard to find :-/
– Marco13
Dec 19 '18 at 22:14

Ah, there it is: stackoverflow.com/q/35296734/3182664 - any objections to close this one as a duplicate of the other?
– Marco13
Dec 19 '18 at 22:16

add a comment |

So I found a curiosity with Java's Map computeIfAbsent (using java8) method and I hope someone can tell me why this is happening as I can not really follow the logic behind that issue.

So, I have a Map with a key (obviously) and the value is a list and I use the computeIfAbsent to create a new list when a key is not set yet. Now when I use an Integer as key I can use the following:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

But when I use a String as key trying to use

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>());

Maybe someone can enlighten me about that. Thanks :)

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

As a side note, Guava's MultiMap already implements what you're looking for.
– chrylis
Dec 19 '18 at 7:28

and its worth mentioning the type of map used in the code for clarity of the question
– nullpointer
Dec 19 '18 at 7:46

I'm pretty sure this is a duplicate, but it's hard to find :-/
– Marco13
Dec 19 '18 at 22:14

Ah, there it is: stackoverflow.com/q/35296734/3182664 - any objections to close this one as a duplicate of the other?
– Marco13
Dec 19 '18 at 22:16

add a comment |

So I found a curiosity with Java's Map computeIfAbsent (using java8) method and I hope someone can tell me why this is happening as I can not really follow the logic behind that issue.

So, I have a Map with a key (obviously) and the value is a list and I use the computeIfAbsent to create a new list when a key is not set yet. Now when I use an Integer as key I can use the following:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

But when I use a String as key trying to use

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>());

Maybe someone can enlighten me about that. Thanks :)

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

So I found a curiosity with Java's Map computeIfAbsent (using java8) method and I hope someone can tell me why this is happening as I can not really follow the logic behind that issue.

So, I have a Map with a key (obviously) and the value is a list and I use the computeIfAbsent to create a new list when a key is not set yet. Now when I use an Integer as key I can use the following:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

But when I use a String as key trying to use

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>());

Maybe someone can enlighten me about that. Thanks :)

java java-8 functional-programming

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

edited Dec 19 '18 at 7:29

asked Dec 19 '18 at 7:27

Fussel

64313

asked Dec 19 '18 at 7:27

Fussel

64313

asked Dec 19 '18 at 7:27

Fussel

64313

As a side note, Guava's MultiMap already implements what you're looking for.
– chrylis
Dec 19 '18 at 7:28

and its worth mentioning the type of map used in the code for clarity of the question
– nullpointer
Dec 19 '18 at 7:46

I'm pretty sure this is a duplicate, but it's hard to find :-/
– Marco13
Dec 19 '18 at 22:14

Ah, there it is: stackoverflow.com/q/35296734/3182664 - any objections to close this one as a duplicate of the other?
– Marco13
Dec 19 '18 at 22:16

add a comment |

As a side note, Guava's MultiMap already implements what you're looking for.
– chrylis
Dec 19 '18 at 7:28

and its worth mentioning the type of map used in the code for clarity of the question
– nullpointer
Dec 19 '18 at 7:46

I'm pretty sure this is a duplicate, but it's hard to find :-/
– Marco13
Dec 19 '18 at 22:14

Ah, there it is: stackoverflow.com/q/35296734/3182664 - any objections to close this one as a duplicate of the other?
– Marco13
Dec 19 '18 at 22:16

As a side note, Guava's MultiMap already implements what you're looking for.
– chrylis
Dec 19 '18 at 7:28

and its worth mentioning the type of map used in the code for clarity of the question
– nullpointer
Dec 19 '18 at 7:46

I'm pretty sure this is a duplicate, but it's hard to find :-/
– Marco13
Dec 19 '18 at 22:14

Ah, there it is: stackoverflow.com/q/35296734/3182664 - any objections to close this one as a duplicate of the other?
– Marco13
Dec 19 '18 at 22:16

add a comment |

2 Answers
2

active

oldest

votes

The mapping function - Function<? super K, ? extends V> mappingFunction - maps a key to a value, so when the key is Integer, ArrayList::new works, since ArrayList has a constructor that takes an int (the initial capacity). On the other hand, it doesn't have a constructor that takes a String.

Since the key probably shouldn't affect the initial capacity of the ArrayList, you should not use a method reference here (in both cases). Use lambda expression.

To make it clearer:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent(1, k -> new ArrayList<>(k));

so it will create an ArrayList with initial capacity of 1.

On the other hand:

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

where k is a String, so it doesn't pass compilation.

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

1

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

2

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

|
show 2 more comments

Your second attempt

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

would actually be equal to

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

and since ArrayList has no constructor taking a String as a parameter, it doesn't work. The first example works as it creates a list with one element.

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

add a comment |

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2 Answers
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active

oldest

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2 Answers
2

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oldest

votes

Since the key probably shouldn't affect the initial capacity of the ArrayList, you should not use a method reference here (in both cases). Use lambda expression.

To make it clearer:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent(1, k -> new ArrayList<>(k));

so it will create an ArrayList with initial capacity of 1.

On the other hand:

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

where k is a String, so it doesn't pass compilation.

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

1

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

2

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

|
show 2 more comments

Since the key probably shouldn't affect the initial capacity of the ArrayList, you should not use a method reference here (in both cases). Use lambda expression.

To make it clearer:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent(1, k -> new ArrayList<>(k));

so it will create an ArrayList with initial capacity of 1.

On the other hand:

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

where k is a String, so it doesn't pass compilation.

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

1

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

2

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

|
show 2 more comments

Since the key probably shouldn't affect the initial capacity of the ArrayList, you should not use a method reference here (in both cases). Use lambda expression.

To make it clearer:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent(1, k -> new ArrayList<>(k));

so it will create an ArrayList with initial capacity of 1.

On the other hand:

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

where k is a String, so it doesn't pass compilation.

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

Since the key probably shouldn't affect the initial capacity of the ArrayList, you should not use a method reference here (in both cases). Use lambda expression.

To make it clearer:

List<Object> list = map.computeIfAbsent(1, ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent(1, k -> new ArrayList<>(k));

so it will create an ArrayList with initial capacity of 1.

On the other hand:

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

behaves similar to:

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

where k is a String, so it doesn't pass compilation.

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

edited Dec 19 '18 at 7:36

answered Dec 19 '18 at 7:29

Eran

279k37452537

answered Dec 19 '18 at 7:29

Eran

279k37452537

answered Dec 19 '18 at 7:29

Eran

279k37452537

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

1

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

2

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

|
show 2 more comments

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

1

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

2

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

And more specifically, 1 -> ArrayList::new does not in fact produce a list with the value 1 inserted.
– chrylis
Dec 19 '18 at 7:30

So when using ArrayList::new the given key is inserted into the list? That in my example I'd have a List with the value 1 already in it?
– Fussel
Dec 19 '18 at 7:33

@chrylis did the OP expect such behavior (1 being inserted into the List)? I didn't see that mentioned in the question.
– Eran
Dec 19 '18 at 7:33

@Fussel the key will not be part of the list either way, but you are correct to use lambda expressions in both cases (Integer and String key). No, the List being List<String> makes no difference, since the key is not added to the List, it is passed to the ArrayList constructor (which doesn't work for Strings, since ArrayList has no constructor that accepts a String).
– Eran
Dec 19 '18 at 7:45

@Fussel If you replace 1 with 1 billion you might get a memory error since Java will try to allocate an array that has the capacity of storing 1 billion elements. Note that capacity != actual elements. An empty ArrayList can consume an arbitrary amount of memory depending on the capacity.
– Giacomo Alzetta
Dec 19 '18 at 15:08

|
show 2 more comments

Your second attempt

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

would actually be equal to

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

and since ArrayList has no constructor taking a String as a parameter, it doesn't work. The first example works as it creates a list with one element.

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

add a comment |

Your second attempt

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

would actually be equal to

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

and since ArrayList has no constructor taking a String as a parameter, it doesn't work. The first example works as it creates a list with one element.

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

add a comment |

Your second attempt

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

would actually be equal to

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

and since ArrayList has no constructor taking a String as a parameter, it doesn't work. The first example works as it creates a list with one element.

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

Your second attempt

List<Object> list = map.computeIfAbsent("key", ArrayList::new);

would actually be equal to

List<Object> list = map.computeIfAbsent("key", k -> new ArrayList<>(k));

and since ArrayList has no constructor taking a String as a parameter, it doesn't work. The first example works as it creates a list with one element.

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

answered Dec 19 '18 at 7:30

Jan Gassen

1,69011531

add a comment |

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