Why java “putAll” cannot deep copy Map's value elements?
I got this code snippet:
public static void main(String args)
Map<String, Set<String>> map = new HashMap<>();
Set<String> set = new HashSet<>();
set.add("user1");
set.add("user2");
map.put("key1", set);
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);// I expect all elements are copied
map.get("key1").add("user3");// add 1 element in "map"
System.out.println(map2.get("key1").size()); // "map2" was affected
In fact the modification of map's set element affected map2, so the program prints "3" instead of "2"
This is weird, I expect that, as long as I used "putAll" method for the new map2 construction, I think both key and value should be deeply cloned?
How to fix my program and make sure map2 is completed divided from map, while copying all elements from map?
Thanks
java dictionary set copy clone
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
add a comment |
I got this code snippet:
public static void main(String args)
Map<String, Set<String>> map = new HashMap<>();
Set<String> set = new HashSet<>();
set.add("user1");
set.add("user2");
map.put("key1", set);
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);// I expect all elements are copied
map.get("key1").add("user3");// add 1 element in "map"
System.out.println(map2.get("key1").size()); // "map2" was affected
In fact the modification of map's set element affected map2, so the program prints "3" instead of "2"
This is weird, I expect that, as long as I used "putAll" method for the new map2 construction, I think both key and value should be deeply cloned?
How to fix my program and make sure map2 is completed divided from map, while copying all elements from map?
Thanks
java dictionary set copy clone
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
add a comment |
I got this code snippet:
public static void main(String args)
Map<String, Set<String>> map = new HashMap<>();
Set<String> set = new HashSet<>();
set.add("user1");
set.add("user2");
map.put("key1", set);
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);// I expect all elements are copied
map.get("key1").add("user3");// add 1 element in "map"
System.out.println(map2.get("key1").size()); // "map2" was affected
In fact the modification of map's set element affected map2, so the program prints "3" instead of "2"
This is weird, I expect that, as long as I used "putAll" method for the new map2 construction, I think both key and value should be deeply cloned?
How to fix my program and make sure map2 is completed divided from map, while copying all elements from map?
Thanks
java dictionary set copy clone
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
I got this code snippet:
public static void main(String args)
Map<String, Set<String>> map = new HashMap<>();
Set<String> set = new HashSet<>();
set.add("user1");
set.add("user2");
map.put("key1", set);
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);// I expect all elements are copied
map.get("key1").add("user3");// add 1 element in "map"
System.out.println(map2.get("key1").size()); // "map2" was affected
In fact the modification of map's set element affected map2, so the program prints "3" instead of "2"
This is weird, I expect that, as long as I used "putAll" method for the new map2 construction, I think both key and value should be deeply cloned?
How to fix my program and make sure map2 is completed divided from map, while copying all elements from map?
Thanks
java dictionary set copy clone
java dictionary set copy clone
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
asked Nov 13 '18 at 11:25
Hind ForsumHind Forsum
2,91631741
2,91631741
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
putAll copies references of the keys and values. It does not make copies of the instances referenced by those references.
You'll have to loop (or stream) over the original Map and create copies of all the value Sets:
Map<String, Set<String>> map2 =
map.entrySet()
.stream()
.collect(Collectors.toMap(Map.Entry::getKey,e-> new HashSet<>(e.getValue())));
Note that there's no need to create copies of the keys, since String is immutable.
answered Nov 13 '18 at 11:27
EranEran
284k37462550
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
add a comment |
Another way:
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);
map2.replaceAll((k, v) -> new HashSet<>(v));
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
putAll copies references of the keys and values. It does not make copies of the instances referenced by those references.
You'll have to loop (or stream) over the original Map and create copies of all the value Sets:
Map<String, Set<String>> map2 =
map.entrySet()
.stream()
.collect(Collectors.toMap(Map.Entry::getKey,e-> new HashSet<>(e.getValue())));
Note that there's no need to create copies of the keys, since String is immutable.
answered Nov 13 '18 at 11:27
EranEran
284k37462550
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
add a comment |
putAll copies references of the keys and values. It does not make copies of the instances referenced by those references.
You'll have to loop (or stream) over the original Map and create copies of all the value Sets:
Map<String, Set<String>> map2 =
map.entrySet()
.stream()
.collect(Collectors.toMap(Map.Entry::getKey,e-> new HashSet<>(e.getValue())));
Note that there's no need to create copies of the keys, since String is immutable.
answered Nov 13 '18 at 11:27
EranEran
284k37462550
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
add a comment |
putAll copies references of the keys and values. It does not make copies of the instances referenced by those references.
You'll have to loop (or stream) over the original Map and create copies of all the value Sets:
Map<String, Set<String>> map2 =
map.entrySet()
.stream()
.collect(Collectors.toMap(Map.Entry::getKey,e-> new HashSet<>(e.getValue())));
Note that there's no need to create copies of the keys, since String is immutable.
answered Nov 13 '18 at 11:27
EranEran
284k37462550
putAll copies references of the keys and values. It does not make copies of the instances referenced by those references.
You'll have to loop (or stream) over the original Map and create copies of all the value Sets:
Map<String, Set<String>> map2 =
map.entrySet()
.stream()
.collect(Collectors.toMap(Map.Entry::getKey,e-> new HashSet<>(e.getValue())));
Note that there's no need to create copies of the keys, since String is immutable.
answered Nov 13 '18 at 11:27
EranEran
284k37462550
answered Nov 13 '18 at 11:27
EranEran
284k37462550
answered Nov 13 '18 at 11:27
EranEran
284k37462550
answered Nov 13 '18 at 11:27
EranEran
284k37462550
284k37462550
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
add a comment |
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
That works, thanks!
– Hind Forsum
Nov 13 '18 at 11:30
add a comment |
Another way:
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);
map2.replaceAll((k, v) -> new HashSet<>(v));
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
add a comment |
Another way:
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);
map2.replaceAll((k, v) -> new HashSet<>(v));
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
add a comment |
Another way:
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);
map2.replaceAll((k, v) -> new HashSet<>(v));
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
Another way:
Map<String, Set<String>> map2 = new HashMap<>();
map2.putAll(map);
map2.replaceAll((k, v) -> new HashSet<>(v));
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
answered Nov 13 '18 at 14:21
Federico Peralta SchaffnerFederico Peralta Schaffner
23.1k43676
23.1k43676
add a comment |
add a comment |
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