How to bind a Java Supplier to an instance of an object?

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6
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How can I bind a Java Supplier to an existing instance of an Object? For example, if I want to write my own compareTo() method with this header:

public static int myCompareTo(Object o1, Object o2, Supplier<Comparable> supplier) ...

I want be able to call it like:

myCompareTo("Hello", "Hello2", String::length);

where String (with the capital letter) is a class and no object. So how can I bind the instance o1 to the supplier?

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

add a commentÂ |Â

up vote
6
down vote

favorite

How can I bind a Java Supplier to an existing instance of an Object? For example, if I want to write my own compareTo() method with this header:

public static int myCompareTo(Object o1, Object o2, Supplier<Comparable> supplier) ...

I want be able to call it like:

myCompareTo("Hello", "Hello2", String::length);

where String (with the capital letter) is a class and no object. So how can I bind the instance o1 to the supplier?

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

add a commentÂ |Â

up vote
6
down vote

favorite

How can I bind a Java Supplier to an existing instance of an Object? For example, if I want to write my own compareTo() method with this header:

public static int myCompareTo(Object o1, Object o2, Supplier<Comparable> supplier) ...

I want be able to call it like:

myCompareTo("Hello", "Hello2", String::length);

where String (with the capital letter) is a class and no object. So how can I bind the instance o1 to the supplier?

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

How can I bind a Java Supplier to an existing instance of an Object? For example, if I want to write my own compareTo() method with this header:

public static int myCompareTo(Object o1, Object o2, Supplier<Comparable> supplier) ...

I want be able to call it like:

myCompareTo("Hello", "Hello2", String::length);

where String (with the capital letter) is a class and no object. So how can I bind the instance o1 to the supplier?

java method-reference

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

edited 2 hours ago

LAD

1,635719

edited 2 hours ago

LAD

1,635719

edited 2 hours ago

LAD

1,635719

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

asked 3 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

Philipp NiedergesÃ¤ÃŸ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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4 Answers
4

active

oldest

votes

up vote
1
down vote

You can use

Comparator.comparing(String::length);

to obtain a comparator instance which you can pass to the method.

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

1

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

1

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

add a commentÂ |Â

up vote
1
down vote

Thanks for your answers. Actually I figured it out now. I wanted to have the supplied object instances (o1 and o2) to execute the given method. I found out that Supplier was the wrong interface instead I had to use Function. Here you can see my working simplified example:

public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) 
 return getter.apply(o1).compareTo(getter.apply(o2));

The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.

For example, if you define the method reference as:

Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();

then the equivalent lambda expression being executed is:

(instance) -> instance.referencedMethod()

Additional Information:

Edit: I know I could have done the same by using Comparator, but this example is very simplified. In my application a Function of this kind is neccessary. I had to create a compareTo function that sorts an ArrayList by more than one attribute because the main sorting attribute may not be unique in the list. I want to share my code with you, because I think it can be a interesting insight for you.

public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) 
 for (Function<T, Comparable> getter : getters) 
 int result = getter.apply(o1).compareTo(getter.apply(o2));
 if (result != 0) return result;
 
 return 0;

With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

add a commentÂ |Â

up vote
1
down vote

Actually a more correct way to define your method would be:

private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) 
 return Comparator.comparing(fu).compare(left, right);

answered 15 mins ago

Eugene

62.3k986144

add a commentÂ |Â

up vote
0
down vote

Here's what you were searching for (I believe):

public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) 
 return mapper.apply(o1).compareTo(mapper.apply(o2));

You can call that like so:

compare("str1", "str2", String::length); // 0

answered 3 hours ago

steffen

8,28012052

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
1
down vote

You can use

Comparator.comparing(String::length);

to obtain a comparator instance which you can pass to the method.

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

1

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

1

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

add a commentÂ |Â

up vote
1
down vote

You can use

Comparator.comparing(String::length);

to obtain a comparator instance which you can pass to the method.

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

1

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

1

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

add a commentÂ |Â

up vote
1
down vote

You can use

Comparator.comparing(String::length);

to obtain a comparator instance which you can pass to the method.

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

You can use

Comparator.comparing(String::length);

to obtain a comparator instance which you can pass to the method.

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

edited 2 hours ago

answered 3 hours ago

senjin.hajrulahovic

422412

answered 3 hours ago

senjin.hajrulahovic

422412

answered 3 hours ago

senjin.hajrulahovic

422412

1

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

1

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

add a commentÂ |Â

1

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

1

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

Comparator.comparing(String::length).compare("Hello1", "Hello2"); will be shorter :)
â€“Â Oleksandr
3 hours ago

Of course it will. I tried to satisfy the provided myCompareTo interface.
â€“Â senjin.hajrulahovic
3 hours ago

@Oleksandr should have made that an answer IMO, if you didn't I did
â€“Â Eugene
13 mins ago

add a commentÂ |Â

up vote
1
down vote

public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) 
 return getter.apply(o1).compareTo(getter.apply(o2));

The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.

For example, if you define the method reference as:

Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();

then the equivalent lambda expression being executed is:

(instance) -> instance.referencedMethod()

Additional Information:

public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) 
 for (Function<T, Comparable> getter : getters) 
 int result = getter.apply(o1).compareTo(getter.apply(o2));
 if (result != 0) return result;
 
 return 0;

With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

add a commentÂ |Â

up vote
1
down vote

public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) 
 return getter.apply(o1).compareTo(getter.apply(o2));

The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.

For example, if you define the method reference as:

Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();

then the equivalent lambda expression being executed is:

(instance) -> instance.referencedMethod()

Additional Information:

public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) 
 for (Function<T, Comparable> getter : getters) 
 int result = getter.apply(o1).compareTo(getter.apply(o2));
 if (result != 0) return result;
 
 return 0;

With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

add a commentÂ |Â

up vote
1
down vote

public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) 
 return getter.apply(o1).compareTo(getter.apply(o2));

The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.

For example, if you define the method reference as:

Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();

then the equivalent lambda expression being executed is:

(instance) -> instance.referencedMethod()

Additional Information:

public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) 
 for (Function<T, Comparable> getter : getters) 
 int result = getter.apply(o1).compareTo(getter.apply(o2));
 if (result != 0) return result;
 
 return 0;

With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

public static <T> int myCompareTo(T o1, T o2, Function<T, Comparable> getter) 
 return getter.apply(o1).compareTo(getter.apply(o2));

The reason, the interface has to be Function and not Supplier is, that only Function is equivalent to a lambda expression taking an object and calls the referenced method on the object.

For example, if you define the method reference as:

Function<TypeOfInstance, ReturnTypeOfReferencedMethod> methodReference = TypeOfInstance::referencedMethod();

then the equivalent lambda expression being executed is:

(instance) -> instance.referencedMethod()

Additional Information:

public static <T> int ultimateCompare(T o1, T o2, Function<T, Comparable>... getters) 
 for (Function<T, Comparable> getter : getters) 
 int result = getter.apply(o1).compareTo(getter.apply(o2));
 if (result != 0) return result;
 
 return 0;

With this for example, you can sort a list of persons by last name and if two of them are identical, you can use the first name to sort. With this solution you can change sorting at runtime.

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

edited 1 hour ago

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

answered 2 hours ago

Philipp NiedergesÃ¤ÃŸ

516

New contributor

Philipp NiedergesÃ¤ÃŸ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a commentÂ |Â

up vote
1
down vote

Actually a more correct way to define your method would be:

private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) 
 return Comparator.comparing(fu).compare(left, right);

answered 15 mins ago

Eugene

62.3k986144

add a commentÂ |Â

up vote
1
down vote

Actually a more correct way to define your method would be:

private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) 
 return Comparator.comparing(fu).compare(left, right);

answered 15 mins ago

Eugene

62.3k986144

add a commentÂ |Â

up vote
1
down vote

Actually a more correct way to define your method would be:

private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) 
 return Comparator.comparing(fu).compare(left, right);

answered 15 mins ago

Eugene

62.3k986144

Actually a more correct way to define your method would be:

private static <T, U extends Comparable<? super U>> int myCompareTo(T left, T right, Function<T, U> fu) 
 return Comparator.comparing(fu).compare(left, right);

answered 15 mins ago

Eugene

62.3k986144

answered 15 mins ago

Eugene

62.3k986144

answered 15 mins ago

Eugene

62.3k986144

answered 15 mins ago

Eugene

62.3k986144

add a commentÂ |Â

up vote
0
down vote

Here's what you were searching for (I believe):

public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) 
 return mapper.apply(o1).compareTo(mapper.apply(o2));

You can call that like so:

compare("str1", "str2", String::length); // 0

answered 3 hours ago

steffen

8,28012052

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

add a commentÂ |Â

up vote
0
down vote

Here's what you were searching for (I believe):

public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) 
 return mapper.apply(o1).compareTo(mapper.apply(o2));

You can call that like so:

compare("str1", "str2", String::length); // 0

answered 3 hours ago

steffen

8,28012052

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

add a commentÂ |Â

up vote
0
down vote

Here's what you were searching for (I believe):

public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) 
 return mapper.apply(o1).compareTo(mapper.apply(o2));

You can call that like so:

compare("str1", "str2", String::length); // 0

answered 3 hours ago

steffen

8,28012052

Here's what you were searching for (I believe):

public static <T, U extends Comparable<U>> int compare(T o1, T o2, Function<T, U> mapper) 
 return mapper.apply(o1).compareTo(mapper.apply(o2));

You can call that like so:

compare("str1", "str2", String::length); // 0

answered 3 hours ago

steffen

8,28012052

answered 3 hours ago

steffen

8,28012052

answered 3 hours ago

steffen

8,28012052

answered 3 hours ago

steffen

8,28012052

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

add a commentÂ |Â

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

I think you were the closest, here, but not entirely close
â€“Â Eugene
14 mins ago

add a commentÂ |Â

Philipp NiedergesÃ¤ÃŸ is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Philipp NiedergesÃ¤ÃŸ is a new contributor. Be nice, and check out our Code of Conduct.

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