ClassNotFoundException with Class.forName(“int”)

-2

Please note that I can't get forName() to work specifically with "int".
On the other hand, it works alright with a class I created in the same package as

Class.forName("mypackage.dummyclass")

Is that something to do with fully qualified names? I've tried "java.lang.int" and "java.lang.integer", but it didn't help.

asked Jan 1 at 18:20

vanhemt

647

Try java.lang.Integer, with an upper-case I.

– Joe C
Jan 1 at 18:21

1

I would strongly advice not to do so. What makes you do that?

– Dorian Gray
Jan 1 at 18:23

Class.forName is for class objects, int is primitive data type, Integer is wrapper class

– Deadpool
Jan 1 at 18:23

@DorianGray Just got confused working through somebody else's example that generates classes on the fly. Will change to int.class now that I'm clear as to what's going on.

– vanhemt
Jan 2 at 1:57

add a comment |

-2

Please note that I can't get forName() to work specifically with "int".
On the other hand, it works alright with a class I created in the same package as

Class.forName("mypackage.dummyclass")

Is that something to do with fully qualified names? I've tried "java.lang.int" and "java.lang.integer", but it didn't help.

asked Jan 1 at 18:20

vanhemt

647

Try java.lang.Integer, with an upper-case I.

– Joe C
Jan 1 at 18:21

1

I would strongly advice not to do so. What makes you do that?

– Dorian Gray
Jan 1 at 18:23

Class.forName is for class objects, int is primitive data type, Integer is wrapper class

– Deadpool
Jan 1 at 18:23

@DorianGray Just got confused working through somebody else's example that generates classes on the fly. Will change to int.class now that I'm clear as to what's going on.

– vanhemt
Jan 2 at 1:57

add a comment |

-2

Please note that I can't get forName() to work specifically with "int".
On the other hand, it works alright with a class I created in the same package as

Class.forName("mypackage.dummyclass")

Is that something to do with fully qualified names? I've tried "java.lang.int" and "java.lang.integer", but it didn't help.

asked Jan 1 at 18:20

vanhemt

647

Please note that I can't get forName() to work specifically with "int".
On the other hand, it works alright with a class I created in the same package as

Class.forName("mypackage.dummyclass")

Is that something to do with fully qualified names? I've tried "java.lang.int" and "java.lang.integer", but it didn't help.

java

asked Jan 1 at 18:20

vanhemt

647

asked Jan 1 at 18:20

vanhemt

647

asked Jan 1 at 18:20

vanhemt

647

asked Jan 1 at 18:20

vanhemt

647

asked Jan 1 at 18:20

vanhemt

647

Try java.lang.Integer, with an upper-case I.

– Joe C
Jan 1 at 18:21

1

I would strongly advice not to do so. What makes you do that?

– Dorian Gray
Jan 1 at 18:23

Class.forName is for class objects, int is primitive data type, Integer is wrapper class

– Deadpool
Jan 1 at 18:23

@DorianGray Just got confused working through somebody else's example that generates classes on the fly. Will change to int.class now that I'm clear as to what's going on.

– vanhemt
Jan 2 at 1:57

add a comment |

Try java.lang.Integer, with an upper-case I.

– Joe C
Jan 1 at 18:21

1

I would strongly advice not to do so. What makes you do that?

– Dorian Gray
Jan 1 at 18:23

Class.forName is for class objects, int is primitive data type, Integer is wrapper class

– Deadpool
Jan 1 at 18:23

@DorianGray Just got confused working through somebody else's example that generates classes on the fly. Will change to int.class now that I'm clear as to what's going on.

– vanhemt
Jan 2 at 1:57

Try java.lang.Integer, with an upper-case I.

– Joe C
Jan 1 at 18:21

I would strongly advice not to do so. What makes you do that?

– Dorian Gray
Jan 1 at 18:23

Class.forName is for class objects, int is primitive data type, Integer is wrapper class

– Deadpool
Jan 1 at 18:23

@DorianGray Just got confused working through somebody else's example that generates classes on the fly. Will change to int.class now that I'm clear as to what's going on.

– vanhemt
Jan 2 at 1:57

add a comment |

5 Answers
5

active

oldest

votes

The correct syntax for getting the Class object associated with primitive types, like int, is

int.class

Likewise, you use byte.class, long.class, boolean.class, etc. for the other primitive types.

While there's a special relationship between int and java.lang.Integer, int.class is not the same thing as Integer.class.

Note that you cannot use Class.forName with a primitive type name.

As documented in the Java tutorial:

Class.forName()

If the fully-qualified name of a class is available, it is possible to get the corresponding Class using the static method Class.forName(). This cannot be used for primitive types.

And from the same tutorial, on how to use the .class syntax with primitives:

The .class Syntax

If the type is available but there is no instance then it is possible to obtain a Class by appending ".class" to the name of the type. This is also the easiest way to obtain the Class for a primitive type.

boolean b;
Class c = b.getClass(); // compile-time error
Class c = boolean.class; // correct

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

1

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

1

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

|
show 1 more comment

int is a primitive type in Java, not a class, so try java.lang.Integer instead (which is not exactly the same thing but is interchangeable for many purposes)

answered Jan 1 at 18:23

johnheroy

32615

add a comment |

Class.forName is for class objects but int is primitive data type.

Try java.lang.Integer instead.

answered Jan 1 at 21:30

Alon

608129

add a comment |

int is a primitive data type, not a class. Integer class can be used for int. So you should use "Integer".

answered Jan 1 at 18:25

mahfuj asif

31617

add a comment |

Instead of java.lang.integer, try java.lang.Integer (With Capital I)

The above will work because, Class.forName() tries to create an object of the full class name passed as String arg in it.

java.lang.Integer - This is a class

int - This is not a class. It is a primitive data type. Hence, failing.

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53997837%2fclassnotfoundexception-with-class-fornameint%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

The correct syntax for getting the Class object associated with primitive types, like int, is

int.class

Likewise, you use byte.class, long.class, boolean.class, etc. for the other primitive types.

While there's a special relationship between int and java.lang.Integer, int.class is not the same thing as Integer.class.

Note that you cannot use Class.forName with a primitive type name.

As documented in the Java tutorial:

Class.forName()

If the fully-qualified name of a class is available, it is possible to get the corresponding Class using the static method Class.forName(). This cannot be used for primitive types.

And from the same tutorial, on how to use the .class syntax with primitives:

The .class Syntax

If the type is available but there is no instance then it is possible to obtain a Class by appending ".class" to the name of the type. This is also the easiest way to obtain the Class for a primitive type.

boolean b;
Class c = b.getClass(); // compile-time error
Class c = boolean.class; // correct

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

1

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

1

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

|
show 1 more comment

The correct syntax for getting the Class object associated with primitive types, like int, is

int.class

Likewise, you use byte.class, long.class, boolean.class, etc. for the other primitive types.

While there's a special relationship between int and java.lang.Integer, int.class is not the same thing as Integer.class.

Note that you cannot use Class.forName with a primitive type name.

As documented in the Java tutorial:

Class.forName()

If the fully-qualified name of a class is available, it is possible to get the corresponding Class using the static method Class.forName(). This cannot be used for primitive types.

And from the same tutorial, on how to use the .class syntax with primitives:

The .class Syntax

If the type is available but there is no instance then it is possible to obtain a Class by appending ".class" to the name of the type. This is also the easiest way to obtain the Class for a primitive type.

boolean b;
Class c = b.getClass(); // compile-time error
Class c = boolean.class; // correct

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

1

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

1

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

|
show 1 more comment

The correct syntax for getting the Class object associated with primitive types, like int, is

int.class

Likewise, you use byte.class, long.class, boolean.class, etc. for the other primitive types.

While there's a special relationship between int and java.lang.Integer, int.class is not the same thing as Integer.class.

Note that you cannot use Class.forName with a primitive type name.

As documented in the Java tutorial:

Class.forName()

If the fully-qualified name of a class is available, it is possible to get the corresponding Class using the static method Class.forName(). This cannot be used for primitive types.

And from the same tutorial, on how to use the .class syntax with primitives:

The .class Syntax

If the type is available but there is no instance then it is possible to obtain a Class by appending ".class" to the name of the type. This is also the easiest way to obtain the Class for a primitive type.

boolean b;
Class c = b.getClass(); // compile-time error
Class c = boolean.class; // correct

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

The correct syntax for getting the Class object associated with primitive types, like int, is

int.class

Likewise, you use byte.class, long.class, boolean.class, etc. for the other primitive types.

While there's a special relationship between int and java.lang.Integer, int.class is not the same thing as Integer.class.

Note that you cannot use Class.forName with a primitive type name.

As documented in the Java tutorial:

Class.forName()

If the fully-qualified name of a class is available, it is possible to get the corresponding Class using the static method Class.forName(). This cannot be used for primitive types.

And from the same tutorial, on how to use the .class syntax with primitives:

The .class Syntax

If the type is available but there is no instance then it is possible to obtain a Class by appending ".class" to the name of the type. This is also the easiest way to obtain the Class for a primitive type.

boolean b;
Class c = b.getClass(); // compile-time error
Class c = boolean.class; // correct

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

edited Jan 1 at 20:16

answered Jan 1 at 18:59

ernest_k

23.5k42749

answered Jan 1 at 18:59

ernest_k

23.5k42749

answered Jan 1 at 18:59

ernest_k

23.5k42749

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

1

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

1

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

|
show 1 more comment

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

1

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

1

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

but still i'm not able to access at using Class.forName()

– Deadpool
Jan 1 at 19:57

@Deadpool That's right, you can't use Class.forName("int") for that, that's why Java gave a special syntax for that. int.class is what you're looking for, why do you want to use Class.forname? Are you loading classes from dynamic names/strings?

– ernest_k
Jan 1 at 20:00

Class.forName() must work for all the classes and interfaces, so curious that int has class and it is not working for that, i'm curious to know why? either those classes are written in some other Native languages ?

– Deadpool
Jan 1 at 20:07

That's perfect, thanks for links

– Deadpool
Jan 1 at 20:19

@ernest_k I just thought, since I can do int.class, that same class could be accessed through forName. Hence the confusion.

– vanhemt
Jan 2 at 1:54

|
show 1 more comment

int is a primitive type in Java, not a class, so try java.lang.Integer instead (which is not exactly the same thing but is interchangeable for many purposes)

answered Jan 1 at 18:23

johnheroy

32615

add a comment |

int is a primitive type in Java, not a class, so try java.lang.Integer instead (which is not exactly the same thing but is interchangeable for many purposes)

answered Jan 1 at 18:23

johnheroy

32615

add a comment |

int is a primitive type in Java, not a class, so try java.lang.Integer instead (which is not exactly the same thing but is interchangeable for many purposes)

answered Jan 1 at 18:23

johnheroy

32615

int is a primitive type in Java, not a class, so try java.lang.Integer instead (which is not exactly the same thing but is interchangeable for many purposes)

answered Jan 1 at 18:23

johnheroy

32615

answered Jan 1 at 18:23

johnheroy

32615

answered Jan 1 at 18:23

johnheroy

32615

answered Jan 1 at 18:23

johnheroy

32615

add a comment |

Class.forName is for class objects but int is primitive data type.

Try java.lang.Integer instead.

answered Jan 1 at 21:30

Alon

608129

add a comment |

Class.forName is for class objects but int is primitive data type.

Try java.lang.Integer instead.

answered Jan 1 at 21:30

Alon

608129

add a comment |

Class.forName is for class objects but int is primitive data type.

Try java.lang.Integer instead.

answered Jan 1 at 21:30

Alon

608129

Class.forName is for class objects but int is primitive data type.

Try java.lang.Integer instead.

answered Jan 1 at 21:30

Alon

608129

answered Jan 1 at 21:30

Alon

608129

answered Jan 1 at 21:30

Alon

608129

answered Jan 1 at 21:30

Alon

608129

add a comment |

int is a primitive data type, not a class. Integer class can be used for int. So you should use "Integer".

answered Jan 1 at 18:25

mahfuj asif

31617

add a comment |

int is a primitive data type, not a class. Integer class can be used for int. So you should use "Integer".

answered Jan 1 at 18:25

mahfuj asif

31617

add a comment |

int is a primitive data type, not a class. Integer class can be used for int. So you should use "Integer".

answered Jan 1 at 18:25

mahfuj asif

31617

int is a primitive data type, not a class. Integer class can be used for int. So you should use "Integer".

answered Jan 1 at 18:25

mahfuj asif

31617

answered Jan 1 at 18:25

mahfuj asif

31617

answered Jan 1 at 18:25

mahfuj asif

31617

answered Jan 1 at 18:25

mahfuj asif

31617

add a comment |

Instead of java.lang.integer, try java.lang.Integer (With Capital I)

The above will work because, Class.forName() tries to create an object of the full class name passed as String arg in it.

java.lang.Integer - This is a class

int - This is not a class. It is a primitive data type. Hence, failing.

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

add a comment |

Instead of java.lang.integer, try java.lang.Integer (With Capital I)

The above will work because, Class.forName() tries to create an object of the full class name passed as String arg in it.

java.lang.Integer - This is a class

int - This is not a class. It is a primitive data type. Hence, failing.

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

add a comment |

Instead of java.lang.integer, try java.lang.Integer (With Capital I)

The above will work because, Class.forName() tries to create an object of the full class name passed as String arg in it.

java.lang.Integer - This is a class

int - This is not a class. It is a primitive data type. Hence, failing.

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

Instead of java.lang.integer, try java.lang.Integer (With Capital I)

The above will work because, Class.forName() tries to create an object of the full class name passed as String arg in it.

java.lang.Integer - This is a class

int - This is not a class. It is a primitive data type. Hence, failing.

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

answered Jan 1 at 18:26

Pankaj Singhal

6,54452344

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu