Mixing
Java Generics casts strangely
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
I am using java 8.
I recently came across this:
public class Test
public static void main(String args)
String ss = "" + (Test.<Integer>abc(2));
System.out.println(Test.<Integer>abc(2));
public static <T> T abc(T a)
String s = "adsa";
return (T) s;
This does not throw a java.lang.ClassCastException. Why is that?
I always thought + and System.out.println calls toString. But when I try to do that it throws an Exception as expected.
String sss = (Test.<Integer>abc(2)).toString();
java generics casting
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
10
down vote
favorite
I am using java 8.
I recently came across this:
public class Test
public static void main(String args)
String ss = "" + (Test.<Integer>abc(2));
System.out.println(Test.<Integer>abc(2));
public static <T> T abc(T a)
String s = "adsa";
return (T) s;
This does not throw a java.lang.ClassCastException. Why is that?
I always thought + and System.out.println calls toString. But when I try to do that it throws an Exception as expected.
String sss = (Test.<Integer>abc(2)).toString();
java generics casting
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t knowTisInteger. The compiler translated(T)to(Object)due to the declaration of<T>.
– VGR
30 mins ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
I am using java 8.
I recently came across this:
public class Test
public static void main(String args)
String ss = "" + (Test.<Integer>abc(2));
System.out.println(Test.<Integer>abc(2));
public static <T> T abc(T a)
String s = "adsa";
return (T) s;
This does not throw a java.lang.ClassCastException. Why is that?
I always thought + and System.out.println calls toString. But when I try to do that it throws an Exception as expected.
String sss = (Test.<Integer>abc(2)).toString();
java generics casting
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am using java 8.
I recently came across this:
public class Test
public static void main(String args)
String ss = "" + (Test.<Integer>abc(2));
System.out.println(Test.<Integer>abc(2));
public static <T> T abc(T a)
String s = "adsa";
return (T) s;
This does not throw a java.lang.ClassCastException. Why is that?
I always thought + and System.out.println calls toString. But when I try to do that it throws an Exception as expected.
String sss = (Test.<Integer>abc(2)).toString();
java generics casting
java generics casting
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 35 mins ago
zzrv
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 35 mins ago
zzrv
633
asked 35 mins ago
zzrv
633
633
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
zzrv is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t knowTisInteger. The compiler translated(T)to(Object)due to the declaration of<T>.
– VGR
30 mins ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago
add a comment |Â
5
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t knowTisInteger. The compiler translated(T)to(Object)due to the declaration of<T>.
– VGR
30 mins ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago
5
5
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t know
T is Integer. The compiler translated (T) to (Object) due to the declaration of <T>.– VGR
30 mins ago
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t know
T is Integer. The compiler translated (T) to (Object) due to the declaration of <T>.– VGR
30 mins ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
It doesn't throw a ClassCastException because all generic type information is stripped from the compiled code (a process called type erasure). Basically, any type parameter is replaced by Object. That's why the first version works. It's also why the code compiles at all. If you ask the compiler to warn about unchecked or unsafe operations with the -Xlint:unchecked flag, you'll get a warning about an unchecked cast in the return statement of abc().
With this statement:
String sss = (Test.<Integer>abc(2)).toString();
the story is a bit different. While the type parameter T is replaced by Object, the calling code gets translated into byte code that explicitly casts the result to Integer. It is as if the code were written with a method with signature static Object abc(Object) and the statement were written:
String sss = ((Integer) Test.abc(Integer.valueOf(2))).toString();
That is, not only does the cast inside abc() go away due to type erasure, a new cast is inserted by the compiler in the calling code. This cast generates a ClassCastException at run time because the object returned from abc() is a String, not an Integer.
Note that the statement
String ss = "" + (Test.<Integer>abc(2));
doesn't require a cast because the compiler translates it into something that might have been written:
String ss = "" + String.valueOf(Test.<Integer>abc(2));
The compiler is smart enough to recognize that no cast is needed for this code.
answered 24 mins ago
Ted Hopp
197k40307416
add a comment |Â
up vote
3
down vote
Generics disappear at runtime, so a cast to T is really just a cast to Object (which the compiler will actually just get rid of), so there's no class cast exception.
abc is just a method which takes an object and returns an object. StringBuilder.append(Object) is the method which is called, as can be seen from the bytecode:
...
16: invokestatic #7 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
19: invokevirtual #8 // Method java/lang/StringBuilder.append:(Ljava/lang/Object;)Ljava/lang/StringBuilder;
...
When you do
String sss = (Test.<Integer>abc(2)).toString();
Then the bytecode is
...
4: invokestatic #3 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
7: checkcast #4 // class java/lang/Integer
10: invokevirtual #5 // Method java/lang/Integer.toString:()Ljava/lang/String;
...
Your code fails at the checkcast operation, which was not present before.
answered 21 mins ago
Michael
17.4k73066
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It doesn't throw a ClassCastException because all generic type information is stripped from the compiled code (a process called type erasure). Basically, any type parameter is replaced by Object. That's why the first version works. It's also why the code compiles at all. If you ask the compiler to warn about unchecked or unsafe operations with the -Xlint:unchecked flag, you'll get a warning about an unchecked cast in the return statement of abc().
With this statement:
String sss = (Test.<Integer>abc(2)).toString();
the story is a bit different. While the type parameter T is replaced by Object, the calling code gets translated into byte code that explicitly casts the result to Integer. It is as if the code were written with a method with signature static Object abc(Object) and the statement were written:
String sss = ((Integer) Test.abc(Integer.valueOf(2))).toString();
That is, not only does the cast inside abc() go away due to type erasure, a new cast is inserted by the compiler in the calling code. This cast generates a ClassCastException at run time because the object returned from abc() is a String, not an Integer.
Note that the statement
String ss = "" + (Test.<Integer>abc(2));
doesn't require a cast because the compiler translates it into something that might have been written:
String ss = "" + String.valueOf(Test.<Integer>abc(2));
The compiler is smart enough to recognize that no cast is needed for this code.
answered 24 mins ago
Ted Hopp
197k40307416
add a comment |Â
up vote
4
down vote
It doesn't throw a ClassCastException because all generic type information is stripped from the compiled code (a process called type erasure). Basically, any type parameter is replaced by Object. That's why the first version works. It's also why the code compiles at all. If you ask the compiler to warn about unchecked or unsafe operations with the -Xlint:unchecked flag, you'll get a warning about an unchecked cast in the return statement of abc().
With this statement:
String sss = (Test.<Integer>abc(2)).toString();
the story is a bit different. While the type parameter T is replaced by Object, the calling code gets translated into byte code that explicitly casts the result to Integer. It is as if the code were written with a method with signature static Object abc(Object) and the statement were written:
String sss = ((Integer) Test.abc(Integer.valueOf(2))).toString();
That is, not only does the cast inside abc() go away due to type erasure, a new cast is inserted by the compiler in the calling code. This cast generates a ClassCastException at run time because the object returned from abc() is a String, not an Integer.
Note that the statement
String ss = "" + (Test.<Integer>abc(2));
doesn't require a cast because the compiler translates it into something that might have been written:
String ss = "" + String.valueOf(Test.<Integer>abc(2));
The compiler is smart enough to recognize that no cast is needed for this code.
answered 24 mins ago
Ted Hopp
197k40307416
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It doesn't throw a ClassCastException because all generic type information is stripped from the compiled code (a process called type erasure). Basically, any type parameter is replaced by Object. That's why the first version works. It's also why the code compiles at all. If you ask the compiler to warn about unchecked or unsafe operations with the -Xlint:unchecked flag, you'll get a warning about an unchecked cast in the return statement of abc().
With this statement:
String sss = (Test.<Integer>abc(2)).toString();
the story is a bit different. While the type parameter T is replaced by Object, the calling code gets translated into byte code that explicitly casts the result to Integer. It is as if the code were written with a method with signature static Object abc(Object) and the statement were written:
String sss = ((Integer) Test.abc(Integer.valueOf(2))).toString();
That is, not only does the cast inside abc() go away due to type erasure, a new cast is inserted by the compiler in the calling code. This cast generates a ClassCastException at run time because the object returned from abc() is a String, not an Integer.
Note that the statement
String ss = "" + (Test.<Integer>abc(2));
doesn't require a cast because the compiler translates it into something that might have been written:
String ss = "" + String.valueOf(Test.<Integer>abc(2));
The compiler is smart enough to recognize that no cast is needed for this code.
answered 24 mins ago
Ted Hopp
197k40307416
It doesn't throw a ClassCastException because all generic type information is stripped from the compiled code (a process called type erasure). Basically, any type parameter is replaced by Object. That's why the first version works. It's also why the code compiles at all. If you ask the compiler to warn about unchecked or unsafe operations with the -Xlint:unchecked flag, you'll get a warning about an unchecked cast in the return statement of abc().
With this statement:
String sss = (Test.<Integer>abc(2)).toString();
the story is a bit different. While the type parameter T is replaced by Object, the calling code gets translated into byte code that explicitly casts the result to Integer. It is as if the code were written with a method with signature static Object abc(Object) and the statement were written:
String sss = ((Integer) Test.abc(Integer.valueOf(2))).toString();
That is, not only does the cast inside abc() go away due to type erasure, a new cast is inserted by the compiler in the calling code. This cast generates a ClassCastException at run time because the object returned from abc() is a String, not an Integer.
Note that the statement
String ss = "" + (Test.<Integer>abc(2));
doesn't require a cast because the compiler translates it into something that might have been written:
String ss = "" + String.valueOf(Test.<Integer>abc(2));
The compiler is smart enough to recognize that no cast is needed for this code.
answered 24 mins ago
Ted Hopp
197k40307416
edited 13 mins ago
answered 24 mins ago
Ted Hopp
197k40307416
answered 24 mins ago
Ted Hopp
197k40307416
answered 24 mins ago
Ted Hopp
197k40307416
197k40307416
add a comment |Â
add a comment |Â
up vote
3
down vote
Generics disappear at runtime, so a cast to T is really just a cast to Object (which the compiler will actually just get rid of), so there's no class cast exception.
abc is just a method which takes an object and returns an object. StringBuilder.append(Object) is the method which is called, as can be seen from the bytecode:
...
16: invokestatic #7 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
19: invokevirtual #8 // Method java/lang/StringBuilder.append:(Ljava/lang/Object;)Ljava/lang/StringBuilder;
...
When you do
String sss = (Test.<Integer>abc(2)).toString();
Then the bytecode is
...
4: invokestatic #3 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
7: checkcast #4 // class java/lang/Integer
10: invokevirtual #5 // Method java/lang/Integer.toString:()Ljava/lang/String;
...
Your code fails at the checkcast operation, which was not present before.
answered 21 mins ago
Michael
17.4k73066
add a comment |Â
up vote
3
down vote
Generics disappear at runtime, so a cast to T is really just a cast to Object (which the compiler will actually just get rid of), so there's no class cast exception.
abc is just a method which takes an object and returns an object. StringBuilder.append(Object) is the method which is called, as can be seen from the bytecode:
...
16: invokestatic #7 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
19: invokevirtual #8 // Method java/lang/StringBuilder.append:(Ljava/lang/Object;)Ljava/lang/StringBuilder;
...
When you do
String sss = (Test.<Integer>abc(2)).toString();
Then the bytecode is
...
4: invokestatic #3 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
7: checkcast #4 // class java/lang/Integer
10: invokevirtual #5 // Method java/lang/Integer.toString:()Ljava/lang/String;
...
Your code fails at the checkcast operation, which was not present before.
answered 21 mins ago
Michael
17.4k73066
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Generics disappear at runtime, so a cast to T is really just a cast to Object (which the compiler will actually just get rid of), so there's no class cast exception.
abc is just a method which takes an object and returns an object. StringBuilder.append(Object) is the method which is called, as can be seen from the bytecode:
...
16: invokestatic #7 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
19: invokevirtual #8 // Method java/lang/StringBuilder.append:(Ljava/lang/Object;)Ljava/lang/StringBuilder;
...
When you do
String sss = (Test.<Integer>abc(2)).toString();
Then the bytecode is
...
4: invokestatic #3 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
7: checkcast #4 // class java/lang/Integer
10: invokevirtual #5 // Method java/lang/Integer.toString:()Ljava/lang/String;
...
Your code fails at the checkcast operation, which was not present before.
answered 21 mins ago
Michael
17.4k73066
Generics disappear at runtime, so a cast to T is really just a cast to Object (which the compiler will actually just get rid of), so there's no class cast exception.
abc is just a method which takes an object and returns an object. StringBuilder.append(Object) is the method which is called, as can be seen from the bytecode:
...
16: invokestatic #7 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
19: invokevirtual #8 // Method java/lang/StringBuilder.append:(Ljava/lang/Object;)Ljava/lang/StringBuilder;
...
When you do
String sss = (Test.<Integer>abc(2)).toString();
Then the bytecode is
...
4: invokestatic #3 // Method abc:(Ljava/lang/Object;)Ljava/lang/Object;
7: checkcast #4 // class java/lang/Integer
10: invokevirtual #5 // Method java/lang/Integer.toString:()Ljava/lang/String;
...
Your code fails at the checkcast operation, which was not present before.
answered 21 mins ago
Michael
17.4k73066
edited 4 mins ago
answered 21 mins ago
Michael
17.4k73066
answered 21 mins ago
Michael
17.4k73066
answered 21 mins ago
Michael
17.4k73066
17.4k73066
add a comment |Â
add a comment |Â
zzrv is a new contributor. Be nice, and check out our Code of Conduct.
zzrv is a new contributor. Be nice, and check out our Code of Conduct.
zzrv is a new contributor. Be nice, and check out our Code of Conduct.
zzrv is a new contributor. Be nice, and check out our Code of Conduct.
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5
Enable all compiler warnings, and you will see the answer: your cast is not “safe.†Meaning, it is not doing what you think it’s doing. Specifically, generics are subject to type erasure, meaning they are only a means of enforcing safety at compile time, not runtime. At runtime, the method doesn’t know
TisInteger. The compiler translated(T)to(Object)due to the declaration of<T>.– VGR
30 mins ago
Possible duplicate of Why does this use of Generics not throw a runtime or compile time exception?
– Oleksandr
51 secs ago