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How to iterate over lambda functions in Java
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up vote
12
down vote
favorite
1
I was able to do it in Python and my Python code is:
signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
a = 5
b = 3
for i in signs.keys():
print(signs[i](a,b))
And the output is:
8
2
How do I do this same thing in Java through HashMap?
java python lambda java-8 java-stream
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
asked Aug 25 at 8:39
Shivansh Kuchhal
746
add a comment |Â
up vote
12
down vote
favorite
1
I was able to do it in Python and my Python code is:
signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
a = 5
b = 3
for i in signs.keys():
print(signs[i](a,b))
And the output is:
8
2
How do I do this same thing in Java through HashMap?
java python lambda java-8 java-stream
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
asked Aug 25 at 8:39
Shivansh Kuchhal
746
Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
2
@Eugene Actually, in Pythonlambdas are limited to no statements at all. Only expressions are allowed.
– wizzwizz4
Aug 25 at 15:28
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41
add a comment |Â
up vote
12
down vote
favorite
1
up vote
12
down vote
favorite
1
1
I was able to do it in Python and my Python code is:
signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
a = 5
b = 3
for i in signs.keys():
print(signs[i](a,b))
And the output is:
8
2
How do I do this same thing in Java through HashMap?
java python lambda java-8 java-stream
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
asked Aug 25 at 8:39
Shivansh Kuchhal
746
I was able to do it in Python and my Python code is:
signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
a = 5
b = 3
for i in signs.keys():
print(signs[i](a,b))
And the output is:
8
2
How do I do this same thing in Java through HashMap?
java python lambda java-8 java-stream
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
asked Aug 25 at 8:39
Shivansh Kuchhal
746
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
edited Aug 25 at 20:36
Peter Mortensen
12.9k1983111
12.9k1983111
asked Aug 25 at 8:39
Shivansh Kuchhal
746
asked Aug 25 at 8:39
Shivansh Kuchhal
746
asked Aug 25 at 8:39
Shivansh Kuchhal
746
746
Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
2
@Eugene Actually, in Pythonlambdas are limited to no statements at all. Only expressions are allowed.
– wizzwizz4
Aug 25 at 15:28
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41
add a comment |Â
Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
2
@Eugene Actually, in Pythonlambdas are limited to no statements at all. Only expressions are allowed.
– wizzwizz4
Aug 25 at 15:28
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41
Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
2
2
@Eugene Actually, in Python
lambdas are limited to no statements at all. Only expressions are allowed.– wizzwizz4
Aug 25 at 15:28
@Eugene Actually, in Python
lambdas are limited to no statements at all. Only expressions are allowed.– wizzwizz4
Aug 25 at 15:28
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
21
down vote
accepted
You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));
answered Aug 25 at 8:45
YCF_L
31.5k103273
3
IntBinaryOperatoris probably a better choice...
– Boris the Spider
Aug 25 at 8:47
2
Thank you @BoristheSpider I don't work withIntBinaryOperatorbefore, I will learn about this, I see your answer and find that you already using it +1
– YCF_L
Aug 25 at 8:55
3
It's(int, int) -> intrather thanInteger- so less boxing. Your answer is good though - +1!
– Boris the Spider
Aug 25 at 8:56
1
If you're using java 10, it's OK to define the map asvar signs = Map.of("+", add, "-", sub);
– Federico Peralta Schaffner
Aug 25 at 22:31
2
Even with boxed values, you can useBinaryOperator<Integer>here, which is a subtype ofBiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions:Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to useIntBinaryOperatorinstead, it’s preferable, as it avoids boxing overhead (and is even shorter thanBinaryOperator<Integer>)…
– Holger
Aug 27 at 8:30
 |Â
show 2 more comments
up vote
14
down vote
Create yourself a nice, typesafe, enum:
enum Operator implements IntBinaryOperator
PLUS("+", Integer::sum),
MINUS("-", (a, b) -> a - b);
private final String symbol;
private final IntBinaryOperator op;
Operator(final String symbol, final IntBinaryOperator op)
this.symbol = symbol;
this.op = op;
public String symbol()
return symbol;
@Override
public int applyAsInt(final int left, final int right)
return op.applyAsInt(left, right);
You may want a lambda that returns double rather than int for other operators.
Now, simply dump that into a Map:
final var operators = Arrays.stream(Operator.values())
.collect(toMap(Operator::symbol, identity()));
For your example though, you don't need a Map at all:
Arrays.stream(Operator.values())
.mapToInt(op -> op.applyAsInt(a,b))
.forEach(System.out::println);
Using:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.toMap;
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
21
down vote
accepted
You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));
answered Aug 25 at 8:45
YCF_L
31.5k103273
3
IntBinaryOperatoris probably a better choice...
– Boris the Spider
Aug 25 at 8:47
2
Thank you @BoristheSpider I don't work withIntBinaryOperatorbefore, I will learn about this, I see your answer and find that you already using it +1
– YCF_L
Aug 25 at 8:55
3
It's(int, int) -> intrather thanInteger- so less boxing. Your answer is good though - +1!
– Boris the Spider
Aug 25 at 8:56
1
If you're using java 10, it's OK to define the map asvar signs = Map.of("+", add, "-", sub);
– Federico Peralta Schaffner
Aug 25 at 22:31
2
Even with boxed values, you can useBinaryOperator<Integer>here, which is a subtype ofBiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions:Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to useIntBinaryOperatorinstead, it’s preferable, as it avoids boxing overhead (and is even shorter thanBinaryOperator<Integer>)…
– Holger
Aug 27 at 8:30
 |Â
show 2 more comments
up vote
21
down vote
accepted
You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));
answered Aug 25 at 8:45
YCF_L
31.5k103273
3
IntBinaryOperatoris probably a better choice...
– Boris the Spider
Aug 25 at 8:47
2
Thank you @BoristheSpider I don't work withIntBinaryOperatorbefore, I will learn about this, I see your answer and find that you already using it +1
– YCF_L
Aug 25 at 8:55
3
It's(int, int) -> intrather thanInteger- so less boxing. Your answer is good though - +1!
– Boris the Spider
Aug 25 at 8:56
1
If you're using java 10, it's OK to define the map asvar signs = Map.of("+", add, "-", sub);
– Federico Peralta Schaffner
Aug 25 at 22:31
2
Even with boxed values, you can useBinaryOperator<Integer>here, which is a subtype ofBiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions:Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to useIntBinaryOperatorinstead, it’s preferable, as it avoids boxing overhead (and is even shorter thanBinaryOperator<Integer>)…
– Holger
Aug 27 at 8:30
 |Â
show 2 more comments
up vote
21
down vote
accepted
up vote
21
down vote
accepted
You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));
answered Aug 25 at 8:45
YCF_L
31.5k103273
You can use BinaryOperator<Integer> in this case like so :
BinaryOperator<Integer> add = (a, b) -> a + b;//lambda a, b : a + b
BinaryOperator<Integer> sub = (a, b) -> a - b;//lambda a, b : a - b
// Then create a new Map which take the sign and the corresponding BinaryOperator
// equivalent to signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, BinaryOperator<Integer>> signs = Map.of("+", add, "-", sub);
int a = 5; // a = 5
int b = 3; // b = 3
// Loop over the sings map and apply the operation
signs.values().forEach(v -> System.out.println(v.apply(a, b)));
Outputs
8
2
Note for Map.of("+", add, "-", sub); I'm using Java 10, If you are not using Java 9+ you can add to your map like so:
Map<String, BinaryOperator<Integer>> signs = new HashMap<>();
signs.put("+", add);
signs.put("-", sub);
Ideone demo
Good practice
As already stated by @Boris the Spider and @Holger in the comments, Its better to use IntBinaryOperator to avoid boxing, in the end your code can look like this :
// signs = "+" : lambda a, b : a + b, "-" : lambda a, b : a - b
Map<String, IntBinaryOperator> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);
int a = 5; // a = 5
int b = 3; // b = 3
// for i in signs.keys(): print(signs[i](a,b))
signs.values().forEach(v -> System.out.println(v.applyAsInt(a, b)));
answered Aug 25 at 8:45
YCF_L
31.5k103273
edited Aug 27 at 8:48
answered Aug 25 at 8:45
YCF_L
31.5k103273
answered Aug 25 at 8:45
YCF_L
31.5k103273
answered Aug 25 at 8:45
YCF_L
31.5k103273
31.5k103273
3
IntBinaryOperatoris probably a better choice...
– Boris the Spider
Aug 25 at 8:47
2
Thank you @BoristheSpider I don't work withIntBinaryOperatorbefore, I will learn about this, I see your answer and find that you already using it +1
– YCF_L
Aug 25 at 8:55
3
It's(int, int) -> intrather thanInteger- so less boxing. Your answer is good though - +1!
– Boris the Spider
Aug 25 at 8:56
1
If you're using java 10, it's OK to define the map asvar signs = Map.of("+", add, "-", sub);
– Federico Peralta Schaffner
Aug 25 at 22:31
2
Even with boxed values, you can useBinaryOperator<Integer>here, which is a subtype ofBiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions:Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to useIntBinaryOperatorinstead, it’s preferable, as it avoids boxing overhead (and is even shorter thanBinaryOperator<Integer>)…
– Holger
Aug 27 at 8:30
 |Â
show 2 more comments
3
IntBinaryOperatoris probably a better choice...
– Boris the Spider
Aug 25 at 8:47
2
Thank you @BoristheSpider I don't work withIntBinaryOperatorbefore, I will learn about this, I see your answer and find that you already using it +1
– YCF_L
Aug 25 at 8:55
3
It's(int, int) -> intrather thanInteger- so less boxing. Your answer is good though - +1!
– Boris the Spider
Aug 25 at 8:56
1
If you're using java 10, it's OK to define the map asvar signs = Map.of("+", add, "-", sub);
– Federico Peralta Schaffner
Aug 25 at 22:31
2
Even with boxed values, you can useBinaryOperator<Integer>here, which is a subtype ofBiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions:Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to useIntBinaryOperatorinstead, it’s preferable, as it avoids boxing overhead (and is even shorter thanBinaryOperator<Integer>)…
– Holger
Aug 27 at 8:30
3
3
IntBinaryOperator is probably a better choice...– Boris the Spider
Aug 25 at 8:47
IntBinaryOperator is probably a better choice...– Boris the Spider
Aug 25 at 8:47
2
2
Thank you @BoristheSpider I don't work with
IntBinaryOperator before, I will learn about this, I see your answer and find that you already using it +1– YCF_L
Aug 25 at 8:55
Thank you @BoristheSpider I don't work with
IntBinaryOperator before, I will learn about this, I see your answer and find that you already using it +1– YCF_L
Aug 25 at 8:55
3
3
It's
(int, int) -> int rather than Integer - so less boxing. Your answer is good though - +1!– Boris the Spider
Aug 25 at 8:56
It's
(int, int) -> int rather than Integer - so less boxing. Your answer is good though - +1!– Boris the Spider
Aug 25 at 8:56
1
1
If you're using java 10, it's OK to define the map as
var signs = Map.of("+", add, "-", sub);– Federico Peralta Schaffner
Aug 25 at 22:31
If you're using java 10, it's OK to define the map as
var signs = Map.of("+", add, "-", sub);– Federico Peralta Schaffner
Aug 25 at 22:31
2
2
Even with boxed values, you can use
BinaryOperator<Integer> here, which is a subtype of BiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions: Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to use IntBinaryOperator instead, it’s preferable, as it avoids boxing overhead (and is even shorter than BinaryOperator<Integer>)…– Holger
Aug 27 at 8:30
Even with boxed values, you can use
BinaryOperator<Integer> here, which is a subtype of BiFunction<Integer, Integer, Integer>. Further, you don’t need to verbosely declare local variables for the functions: Map<String, BinaryOperator<Integer>> signs = Map.of("+", (a, b) -> a + b, "-", (a, b) -> a - b);; this is much closer to the OP’s Python code. Of course, whenever the use case allows to use IntBinaryOperator instead, it’s preferable, as it avoids boxing overhead (and is even shorter than BinaryOperator<Integer>)…– Holger
Aug 27 at 8:30
 |Â
show 2 more comments
up vote
14
down vote
Create yourself a nice, typesafe, enum:
enum Operator implements IntBinaryOperator
PLUS("+", Integer::sum),
MINUS("-", (a, b) -> a - b);
private final String symbol;
private final IntBinaryOperator op;
Operator(final String symbol, final IntBinaryOperator op)
this.symbol = symbol;
this.op = op;
public String symbol()
return symbol;
@Override
public int applyAsInt(final int left, final int right)
return op.applyAsInt(left, right);
You may want a lambda that returns double rather than int for other operators.
Now, simply dump that into a Map:
final var operators = Arrays.stream(Operator.values())
.collect(toMap(Operator::symbol, identity()));
For your example though, you don't need a Map at all:
Arrays.stream(Operator.values())
.mapToInt(op -> op.applyAsInt(a,b))
.forEach(System.out::println);
Using:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.toMap;
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
add a comment |Â
up vote
14
down vote
Create yourself a nice, typesafe, enum:
enum Operator implements IntBinaryOperator
PLUS("+", Integer::sum),
MINUS("-", (a, b) -> a - b);
private final String symbol;
private final IntBinaryOperator op;
Operator(final String symbol, final IntBinaryOperator op)
this.symbol = symbol;
this.op = op;
public String symbol()
return symbol;
@Override
public int applyAsInt(final int left, final int right)
return op.applyAsInt(left, right);
You may want a lambda that returns double rather than int for other operators.
Now, simply dump that into a Map:
final var operators = Arrays.stream(Operator.values())
.collect(toMap(Operator::symbol, identity()));
For your example though, you don't need a Map at all:
Arrays.stream(Operator.values())
.mapToInt(op -> op.applyAsInt(a,b))
.forEach(System.out::println);
Using:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.toMap;
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
add a comment |Â
up vote
14
down vote
up vote
14
down vote
Create yourself a nice, typesafe, enum:
enum Operator implements IntBinaryOperator
PLUS("+", Integer::sum),
MINUS("-", (a, b) -> a - b);
private final String symbol;
private final IntBinaryOperator op;
Operator(final String symbol, final IntBinaryOperator op)
this.symbol = symbol;
this.op = op;
public String symbol()
return symbol;
@Override
public int applyAsInt(final int left, final int right)
return op.applyAsInt(left, right);
You may want a lambda that returns double rather than int for other operators.
Now, simply dump that into a Map:
final var operators = Arrays.stream(Operator.values())
.collect(toMap(Operator::symbol, identity()));
For your example though, you don't need a Map at all:
Arrays.stream(Operator.values())
.mapToInt(op -> op.applyAsInt(a,b))
.forEach(System.out::println);
Using:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.toMap;
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
Create yourself a nice, typesafe, enum:
enum Operator implements IntBinaryOperator
PLUS("+", Integer::sum),
MINUS("-", (a, b) -> a - b);
private final String symbol;
private final IntBinaryOperator op;
Operator(final String symbol, final IntBinaryOperator op)
this.symbol = symbol;
this.op = op;
public String symbol()
return symbol;
@Override
public int applyAsInt(final int left, final int right)
return op.applyAsInt(left, right);
You may want a lambda that returns double rather than int for other operators.
Now, simply dump that into a Map:
final var operators = Arrays.stream(Operator.values())
.collect(toMap(Operator::symbol, identity()));
For your example though, you don't need a Map at all:
Arrays.stream(Operator.values())
.mapToInt(op -> op.applyAsInt(a,b))
.forEach(System.out::println);
Using:
import static java.util.function.Function.identity;
import static java.util.stream.Collectors.toMap;
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
edited Aug 25 at 8:54
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
answered Aug 25 at 8:45
Boris the Spider
43.1k370117
43.1k370117
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Java is only a very tiny bit more verbose
– Walter Tross
Aug 25 at 9:08
except that in python AFAIK lambdas are limited to a single statement, right?
– Eugene
Aug 25 at 12:00
2
@Eugene Actually, in Python
lambdas are limited to no statements at all. Only expressions are allowed.– wizzwizz4
Aug 25 at 15:28
@WalterTross Oops. Values are literals, aren't they? :-/
– wizzwizz4
Aug 25 at 15:31
Literals are values, but not vice versa.
– Midnightas
Aug 25 at 18:41