Java lambda only throwing expression-style instead of statement-style [duplicate]











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  • Why must throw statements be enclosed with a full code block in a lambda body?

    3 answers




In Java (using Java 8 currently), I can write this and all will compile nice and well:



Supplier<Long> asd = () -> {
throw new RuntimeException();
};


Yet, I cannot write this:



Supplier<Long> asd = () -> throw new RuntimeException(); // This won't compile :(


Does anyone know why Java's implementation does not allow such style (expression lambda) and only the statement/code-block style lambda?



I mean, since the lambda is only throwing the RuntimeException, why isn't the JDK able to infer the lambda expression as:



new Supplier<Long>() {
@Override
public Long get() {
throw new RuntimeException();
}
};


Is this documented somewhere in the specs/docs? Is this added only in JDK > 8?










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marked as duplicate by Ridcully, Heinzi, Mark Rotteveel java
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Nov 30 at 15:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
    – JoSSte
    Nov 30 at 12:45






  • 9




    You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
    – Holger
    Nov 30 at 12:48






  • 3




    because throw is not an expression in the same way as return is not an expression perhaps?
    – Joakim Danielson
    Nov 30 at 12:49






  • 1




    Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
    – Heinzi
    Nov 30 at 15:31

















up vote
15
down vote

favorite
1













This question already has an answer here:




  • Why must throw statements be enclosed with a full code block in a lambda body?

    3 answers




In Java (using Java 8 currently), I can write this and all will compile nice and well:



Supplier<Long> asd = () -> {
throw new RuntimeException();
};


Yet, I cannot write this:



Supplier<Long> asd = () -> throw new RuntimeException(); // This won't compile :(


Does anyone know why Java's implementation does not allow such style (expression lambda) and only the statement/code-block style lambda?



I mean, since the lambda is only throwing the RuntimeException, why isn't the JDK able to infer the lambda expression as:



new Supplier<Long>() {
@Override
public Long get() {
throw new RuntimeException();
}
};


Is this documented somewhere in the specs/docs? Is this added only in JDK > 8?










share|improve this question









New contributor




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











marked as duplicate by Ridcully, Heinzi, Mark Rotteveel java
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Nov 30 at 15:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
    – JoSSte
    Nov 30 at 12:45






  • 9




    You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
    – Holger
    Nov 30 at 12:48






  • 3




    because throw is not an expression in the same way as return is not an expression perhaps?
    – Joakim Danielson
    Nov 30 at 12:49






  • 1




    Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
    – Heinzi
    Nov 30 at 15:31















up vote
15
down vote

favorite
1









up vote
15
down vote

favorite
1






1






This question already has an answer here:




  • Why must throw statements be enclosed with a full code block in a lambda body?

    3 answers




In Java (using Java 8 currently), I can write this and all will compile nice and well:



Supplier<Long> asd = () -> {
throw new RuntimeException();
};


Yet, I cannot write this:



Supplier<Long> asd = () -> throw new RuntimeException(); // This won't compile :(


Does anyone know why Java's implementation does not allow such style (expression lambda) and only the statement/code-block style lambda?



I mean, since the lambda is only throwing the RuntimeException, why isn't the JDK able to infer the lambda expression as:



new Supplier<Long>() {
@Override
public Long get() {
throw new RuntimeException();
}
};


Is this documented somewhere in the specs/docs? Is this added only in JDK > 8?










share|improve this question









New contributor




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












This question already has an answer here:




  • Why must throw statements be enclosed with a full code block in a lambda body?

    3 answers




In Java (using Java 8 currently), I can write this and all will compile nice and well:



Supplier<Long> asd = () -> {
throw new RuntimeException();
};


Yet, I cannot write this:



Supplier<Long> asd = () -> throw new RuntimeException(); // This won't compile :(


Does anyone know why Java's implementation does not allow such style (expression lambda) and only the statement/code-block style lambda?



I mean, since the lambda is only throwing the RuntimeException, why isn't the JDK able to infer the lambda expression as:



new Supplier<Long>() {
@Override
public Long get() {
throw new RuntimeException();
}
};


Is this documented somewhere in the specs/docs? Is this added only in JDK > 8?





This question already has an answer here:




  • Why must throw statements be enclosed with a full code block in a lambda body?

    3 answers








java lambda java-8 throw functional-interface






share|improve this question









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











share|improve this question









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Check out our Code of Conduct.









share|improve this question




share|improve this question








edited Nov 30 at 14:14









Jesse de Bruijne

2,45851325




2,45851325






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asked Nov 30 at 12:40









Tamir Nauman

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794




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New contributor





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






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




marked as duplicate by Ridcully, Heinzi, Mark Rotteveel java
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Nov 30 at 15:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Ridcully, Heinzi, Mark Rotteveel java
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Nov 30 at 15:29


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
    – JoSSte
    Nov 30 at 12:45






  • 9




    You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
    – Holger
    Nov 30 at 12:48






  • 3




    because throw is not an expression in the same way as return is not an expression perhaps?
    – Joakim Danielson
    Nov 30 at 12:49






  • 1




    Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
    – Heinzi
    Nov 30 at 15:31




















  • If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
    – JoSSte
    Nov 30 at 12:45






  • 9




    You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
    – Holger
    Nov 30 at 12:48






  • 3




    because throw is not an expression in the same way as return is not an expression perhaps?
    – Joakim Danielson
    Nov 30 at 12:49






  • 1




    Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
    – Heinzi
    Nov 30 at 15:31


















If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
– JoSSte
Nov 30 at 12:45




If I am not mistaken, the {} surrounding your statement throw new RuntimeException() is an "anonymous function". my guess is that the devs decided that all calls should be in a method, and not in the style you are asking for. this is in no way authoritative, hence the comment, rather than an answer...
– JoSSte
Nov 30 at 12:45




9




9




You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
– Holger
Nov 30 at 12:48




You can not use the expression syntax, because a throw statement is not an expression. As simple as that.
– Holger
Nov 30 at 12:48




3




3




because throw is not an expression in the same way as return is not an expression perhaps?
– Joakim Danielson
Nov 30 at 12:49




because throw is not an expression in the same way as return is not an expression perhaps?
– Joakim Danielson
Nov 30 at 12:49




1




1




Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
– Heinzi
Nov 30 at 15:31






Note: C# got throw expressions in version 7. It might be a good idea to propose such a feature for Java as well (especially if Java ever gets a null coalescing operator, because that seems to be the main use case for throw expressions in C#).
– Heinzi
Nov 30 at 15:31














2 Answers
2






active

oldest

votes

















up vote
22
down vote













From the language spec JLS 15.27.2:




A lambda body is either a single expression or a block (§14.2). 




throw is not a single expression (it's a statement); so you have to use it in a block.






share|improve this answer

















  • 1




    Yes, this is a valuable info and a better answer than mine. Have my +1 :)
    – Nikolas
    Nov 30 at 13:13


















up vote
18
down vote













What you wrote is an invalid lambda. There is a difference between the expression with the brackets {} and without. See the example. The following means that 1L is returned.



Supplier<Long> asd = () -> 1L;


which is equivalent to:



Supplier<Long> asd = () -> {
return 1L;
};


However, when you write:



Supplier<Long> asd = () -> throw new RuntimeException();


It would be translated following which is an invalid lambda:



Supplier<Long> asd = () -> {
return throw new RuntimeException(); // invalid expression in Java
};


In a nutshell, you can understand () -> 1L as a shortcut for { return 1L; }.






share|improve this answer





















  • This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
    – Nick
    1 hour ago




















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
22
down vote













From the language spec JLS 15.27.2:




A lambda body is either a single expression or a block (§14.2). 




throw is not a single expression (it's a statement); so you have to use it in a block.






share|improve this answer

















  • 1




    Yes, this is a valuable info and a better answer than mine. Have my +1 :)
    – Nikolas
    Nov 30 at 13:13















up vote
22
down vote













From the language spec JLS 15.27.2:




A lambda body is either a single expression or a block (§14.2). 




throw is not a single expression (it's a statement); so you have to use it in a block.






share|improve this answer

















  • 1




    Yes, this is a valuable info and a better answer than mine. Have my +1 :)
    – Nikolas
    Nov 30 at 13:13













up vote
22
down vote










up vote
22
down vote









From the language spec JLS 15.27.2:




A lambda body is either a single expression or a block (§14.2). 




throw is not a single expression (it's a statement); so you have to use it in a block.






share|improve this answer












From the language spec JLS 15.27.2:




A lambda body is either a single expression or a block (§14.2). 




throw is not a single expression (it's a statement); so you have to use it in a block.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 30 at 12:55









Andy Turner

79.1k878131




79.1k878131








  • 1




    Yes, this is a valuable info and a better answer than mine. Have my +1 :)
    – Nikolas
    Nov 30 at 13:13














  • 1




    Yes, this is a valuable info and a better answer than mine. Have my +1 :)
    – Nikolas
    Nov 30 at 13:13








1




1




Yes, this is a valuable info and a better answer than mine. Have my +1 :)
– Nikolas
Nov 30 at 13:13




Yes, this is a valuable info and a better answer than mine. Have my +1 :)
– Nikolas
Nov 30 at 13:13












up vote
18
down vote













What you wrote is an invalid lambda. There is a difference between the expression with the brackets {} and without. See the example. The following means that 1L is returned.



Supplier<Long> asd = () -> 1L;


which is equivalent to:



Supplier<Long> asd = () -> {
return 1L;
};


However, when you write:



Supplier<Long> asd = () -> throw new RuntimeException();


It would be translated following which is an invalid lambda:



Supplier<Long> asd = () -> {
return throw new RuntimeException(); // invalid expression in Java
};


In a nutshell, you can understand () -> 1L as a shortcut for { return 1L; }.






share|improve this answer





















  • This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
    – Nick
    1 hour ago

















up vote
18
down vote













What you wrote is an invalid lambda. There is a difference between the expression with the brackets {} and without. See the example. The following means that 1L is returned.



Supplier<Long> asd = () -> 1L;


which is equivalent to:



Supplier<Long> asd = () -> {
return 1L;
};


However, when you write:



Supplier<Long> asd = () -> throw new RuntimeException();


It would be translated following which is an invalid lambda:



Supplier<Long> asd = () -> {
return throw new RuntimeException(); // invalid expression in Java
};


In a nutshell, you can understand () -> 1L as a shortcut for { return 1L; }.






share|improve this answer





















  • This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
    – Nick
    1 hour ago















up vote
18
down vote










up vote
18
down vote









What you wrote is an invalid lambda. There is a difference between the expression with the brackets {} and without. See the example. The following means that 1L is returned.



Supplier<Long> asd = () -> 1L;


which is equivalent to:



Supplier<Long> asd = () -> {
return 1L;
};


However, when you write:



Supplier<Long> asd = () -> throw new RuntimeException();


It would be translated following which is an invalid lambda:



Supplier<Long> asd = () -> {
return throw new RuntimeException(); // invalid expression in Java
};


In a nutshell, you can understand () -> 1L as a shortcut for { return 1L; }.






share|improve this answer












What you wrote is an invalid lambda. There is a difference between the expression with the brackets {} and without. See the example. The following means that 1L is returned.



Supplier<Long> asd = () -> 1L;


which is equivalent to:



Supplier<Long> asd = () -> {
return 1L;
};


However, when you write:



Supplier<Long> asd = () -> throw new RuntimeException();


It would be translated following which is an invalid lambda:



Supplier<Long> asd = () -> {
return throw new RuntimeException(); // invalid expression in Java
};


In a nutshell, you can understand () -> 1L as a shortcut for { return 1L; }.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 30 at 12:47









Nikolas

12k53262




12k53262












  • This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
    – Nick
    1 hour ago




















  • This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
    – Nick
    1 hour ago


















This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
– Nick
1 hour ago






This is not entirely the case, since the following program is valid: Consumer<String> consumer = str -> System.out.println(str). This is not elaborated to Consumer<String> consumer = str -> { return System.out.println(str); } (which would similarly be an invalid program). Some special rules were added to accommodate expressions of type void; it's just that these same rules weren't added to accommodate throw statements.
– Nick
1 hour ago





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