Why does this use of Generics not throw a runtime or compile time exception?

up vote
6
down vote

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I've got a method in a class that has a return type specified by use of a generic.

public class SomeMain 

 public static void main(String args) 

 Foo<Integer> foo = new Foo<Integer>();
 System.out.println(foo.getFoo()); // Works, prints out "Foo"

 

 public static class Foo<E> 
 public E getFoo() 
 return (E) "Foo";

With the generic return type, I assumed the return in the above example would evaluate to:

return (Integer) "Foo"; // Inconvertible types !!!

Instead a String is returned and printed correctly.

I get a compilation error if I change the call to be:

String fooString = foo.getFoo(); // Compile error, incompatible types found
System.out.println(fooString);

What am I missing to help me understand what's going on here and why the original version didn't result in a compilation error.

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

add a commentÂ |Â

up vote
6
down vote

favorite

I've got a method in a class that has a return type specified by use of a generic.

public class SomeMain 

 public static void main(String args) 

 Foo<Integer> foo = new Foo<Integer>();
 System.out.println(foo.getFoo()); // Works, prints out "Foo"

 

 public static class Foo<E> 
 public E getFoo() 
 return (E) "Foo";

With the generic return type, I assumed the return in the above example would evaluate to:

return (Integer) "Foo"; // Inconvertible types !!!

Instead a String is returned and printed correctly.

I get a compilation error if I change the call to be:

String fooString = foo.getFoo(); // Compile error, incompatible types found
System.out.println(fooString);

What am I missing to help me understand what's going on here and why the original version didn't result in a compilation error.

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

add a commentÂ |Â

up vote
6
down vote

favorite

I've got a method in a class that has a return type specified by use of a generic.

public class SomeMain 

 public static void main(String args) 

 Foo<Integer> foo = new Foo<Integer>();
 System.out.println(foo.getFoo()); // Works, prints out "Foo"

 

 public static class Foo<E> 
 public E getFoo() 
 return (E) "Foo";

With the generic return type, I assumed the return in the above example would evaluate to:

return (Integer) "Foo"; // Inconvertible types !!!

Instead a String is returned and printed correctly.

I get a compilation error if I change the call to be:

String fooString = foo.getFoo(); // Compile error, incompatible types found
System.out.println(fooString);

What am I missing to help me understand what's going on here and why the original version didn't result in a compilation error.

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

I've got a method in a class that has a return type specified by use of a generic.

public class SomeMain 

 public static void main(String args) 

 Foo<Integer> foo = new Foo<Integer>();
 System.out.println(foo.getFoo()); // Works, prints out "Foo"

 

 public static class Foo<E> 
 public E getFoo() 
 return (E) "Foo";

With the generic return type, I assumed the return in the above example would evaluate to:

return (Integer) "Foo"; // Inconvertible types !!!

Instead a String is returned and printed correctly.

I get a compilation error if I change the call to be:

String fooString = foo.getFoo(); // Compile error, incompatible types found
System.out.println(fooString);

What am I missing to help me understand what's going on here and why the original version didn't result in a compilation error.

java generics

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

edited 52 mins ago

asked 1 hour ago

Ray

25.9k958103

asked 1 hour ago

Ray

25.9k958103

asked 1 hour ago

Ray

25.9k958103

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
6
down vote

accepted

This is because overload resolution resolved your println call to println(Object), since there is no println(Integer).

Keep in mind that Java's generics are erased at runtime. And casts like (E) "Foo" are removed, and are moved to call site. Sometimes this is not necessary, so things are casted to the right type only when needed.

In other words, no casts are performed inside getFoo. The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

The cast is a completely unchecked cast.

No run-time action is performed for such a cast.

After erasure, getFoo returns Object. And that gets passed into println(Object), which is perfectly fine.

If I call this method and pass foo.getFoo, I will get an error:

static void f(Integer i) 
 System.out.println(i);

// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

add a commentÂ |Â

up vote
2
down vote

System.out.println does not have an overload that takes Integer. So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object);.

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String args) throws Exception 
 Foo<Integer> foo = new Foo<Integer>();
 print(foo.getFoo()); //Would also fail with a class cast exception

static void print(Integer in) 
 System.out.println(in);

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer>, and foo.getFoo() returns an Integer and the compiler can pick this up.

answered 1 hour ago

ernest_k

14.5k41429

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
6
down vote

accepted

This is because overload resolution resolved your println call to println(Object), since there is no println(Integer).

In other words, no casts are performed inside getFoo. The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

The cast is a completely unchecked cast.

No run-time action is performed for such a cast.

After erasure, getFoo returns Object. And that gets passed into println(Object), which is perfectly fine.

If I call this method and pass foo.getFoo, I will get an error:

static void f(Integer i) 
 System.out.println(i);

// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

add a commentÂ |Â

up vote
6
down vote

accepted

This is because overload resolution resolved your println call to println(Object), since there is no println(Integer).

In other words, no casts are performed inside getFoo. The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

The cast is a completely unchecked cast.

No run-time action is performed for such a cast.

After erasure, getFoo returns Object. And that gets passed into println(Object), which is perfectly fine.

If I call this method and pass foo.getFoo, I will get an error:

static void f(Integer i) 
 System.out.println(i);

// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

add a commentÂ |Â

up vote
6
down vote

accepted

This is because overload resolution resolved your println call to println(Object), since there is no println(Integer).

In other words, no casts are performed inside getFoo. The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

The cast is a completely unchecked cast.

No run-time action is performed for such a cast.

After erasure, getFoo returns Object. And that gets passed into println(Object), which is perfectly fine.

If I call this method and pass foo.getFoo, I will get an error:

static void f(Integer i) 
 System.out.println(i);

// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

This is because overload resolution resolved your println call to println(Object), since there is no println(Integer).

In other words, no casts are performed inside getFoo. The language spec supports this:

Section 5.5.2 Checked Casts and Unchecked Casts

The cast is a completely unchecked cast.

No run-time action is performed for such a cast.

After erasure, getFoo returns Object. And that gets passed into println(Object), which is perfectly fine.

If I call this method and pass foo.getFoo, I will get an error:

static void f(Integer i) 
 System.out.println(i);

// ...
f(foo.getFoo()); // ClassCastException

because this time it needs to be casted.

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

edited 1 hour ago

answered 1 hour ago

Sweeper

55.8k962119

answered 1 hour ago

Sweeper

55.8k962119

answered 1 hour ago

Sweeper

55.8k962119

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

add a commentÂ |Â

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

Think that new rule should be introduced in lint that highlights generics usage with compile-time constants
â€“Â Maxim
1 hour ago

Interesting. So, the E in the method declaration public E getFoo() does nothing at run-time, only acts a compile time checking on uses of the signature, not what's going in inside the method.
â€“Â Ray
55 mins ago

add a commentÂ |Â

up vote
2
down vote

System.out.println does not have an overload that takes Integer. So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object);.

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String args) throws Exception 
 Foo<Integer> foo = new Foo<Integer>();
 print(foo.getFoo()); //Would also fail with a class cast exception

static void print(Integer in) 
 System.out.println(in);

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer>, and foo.getFoo() returns an Integer and the compiler can pick this up.

answered 1 hour ago

ernest_k

14.5k41429

add a commentÂ |Â

up vote
2
down vote

System.out.println does not have an overload that takes Integer. So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object);.

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String args) throws Exception 
 Foo<Integer> foo = new Foo<Integer>();
 print(foo.getFoo()); //Would also fail with a class cast exception

static void print(Integer in) 
 System.out.println(in);

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer>, and foo.getFoo() returns an Integer and the compiler can pick this up.

answered 1 hour ago

ernest_k

14.5k41429

add a commentÂ |Â

up vote
2
down vote

System.out.println does not have an overload that takes Integer. So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object);.

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String args) throws Exception 
 Foo<Integer> foo = new Foo<Integer>();
 print(foo.getFoo()); //Would also fail with a class cast exception

static void print(Integer in) 
 System.out.println(in);

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer>, and foo.getFoo() returns an Integer and the compiler can pick this up.

answered 1 hour ago

ernest_k

14.5k41429

System.out.println does not have an overload that takes Integer. So this statement:

System.out.println(foo.getFoo());

Is calling System.out.println(Object);.

To verify that it would otherwise fail, try:

Foo<Integer> foo = new Foo<Integer>();
Integer fooInt = foo.getFoo(); //class cast exception

The following will fail in the same way:

public static void main(String args) throws Exception 
 Foo<Integer> foo = new Foo<Integer>();
 print(foo.getFoo()); //Would also fail with a class cast exception

static void print(Integer in) 
 System.out.println(in);

And this is failing compilation for obvious reasons:

String fooString = foo.getFoo(); //can't work

foo is Foo<Integer>, and foo.getFoo() returns an Integer and the compiler can pick this up.

answered 1 hour ago

ernest_k

14.5k41429

answered 1 hour ago

ernest_k

14.5k41429

answered 1 hour ago

ernest_k

14.5k41429

answered 1 hour ago

ernest_k

14.5k41429

add a commentÂ |Â

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