Mixing
Comparing Array sizes of different types in java [duplicate]
This question already has an answer here:
How can I make a method take an array of any type as a parameter?
4 answers
So I want to create a method that validates that two Arrays are of same length like:
validateSameSize(Object first, Object second) {
if (first.length != second.length) throw new Exception();
}
The problem is that this method only works for non-primitive Arrays. If I want to compare a char-Array with another array this does not work. Is there a way to implement this function without too much overhead?
I have already tried
<T,V> validateSameSize(T first, V second)
but since generics also need a class and don't work with primitive types, this does not work. Also
validateSameSize(Array first, Array second)
doesn't work either
java arrays
asked Jan 26 at 18:34
SchottkySchottky
362
marked as duplicate by Sotirios Delimanolis java
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Jan 26 at 20:29
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.
add a comment |
This question already has an answer here:
How can I make a method take an array of any type as a parameter?
4 answers
So I want to create a method that validates that two Arrays are of same length like:
validateSameSize(Object first, Object second) {
if (first.length != second.length) throw new Exception();
}
The problem is that this method only works for non-primitive Arrays. If I want to compare a char-Array with another array this does not work. Is there a way to implement this function without too much overhead?
I have already tried
<T,V> validateSameSize(T first, V second)
but since generics also need a class and don't work with primitive types, this does not work. Also
validateSameSize(Array first, Array second)
doesn't work either
java arrays
asked Jan 26 at 18:34
SchottkySchottky
362
marked as duplicate by Sotirios Delimanolis java
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Jan 26 at 20:29
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.
add a comment |
This question already has an answer here:
How can I make a method take an array of any type as a parameter?
4 answers
So I want to create a method that validates that two Arrays are of same length like:
validateSameSize(Object first, Object second) {
if (first.length != second.length) throw new Exception();
}
The problem is that this method only works for non-primitive Arrays. If I want to compare a char-Array with another array this does not work. Is there a way to implement this function without too much overhead?
I have already tried
<T,V> validateSameSize(T first, V second)
but since generics also need a class and don't work with primitive types, this does not work. Also
validateSameSize(Array first, Array second)
doesn't work either
java arrays
asked Jan 26 at 18:34
SchottkySchottky
362
This question already has an answer here:
How can I make a method take an array of any type as a parameter?
4 answers
So I want to create a method that validates that two Arrays are of same length like:
validateSameSize(Object first, Object second) {
if (first.length != second.length) throw new Exception();
}
The problem is that this method only works for non-primitive Arrays. If I want to compare a char-Array with another array this does not work. Is there a way to implement this function without too much overhead?
I have already tried
<T,V> validateSameSize(T first, V second)
but since generics also need a class and don't work with primitive types, this does not work. Also
validateSameSize(Array first, Array second)
doesn't work either
This question already has an answer here:
How can I make a method take an array of any type as a parameter?
4 answers
java arrays
java arrays
asked Jan 26 at 18:34
SchottkySchottky
362
asked Jan 26 at 18:34
SchottkySchottky
362
asked Jan 26 at 18:34
SchottkySchottky
362
asked Jan 26 at 18:34
SchottkySchottky
362
asked Jan 26 at 18:34
SchottkySchottky
362
362
marked as duplicate by Sotirios Delimanolis java
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Jan 26 at 20:29
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.
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Jan 26 at 20:29
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.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can use Array#getLength:
public static boolean sameSize(Object arrayA, Object arrayB) {
return Array.getLength(arrayA) == Array.getLength(arrayB);
}
It will work with non-primitive arrays as well as with primitive ones:
System.out.println(sameSize(new int[0], new int[100])); // false
System.out.println(sameSize(new char[0], new int[0])); // true
System.out.println(sameSize(new Object[0], new Object[0])); // true
System.out.println(sameSize(new Object[0], new List[0])); // true
Also don't forget that passing an Object that is not an array to Array#getLength will result in IllegalArgumentException.
This code:
Object notArray = 100;
System.out.println(Array.getLength(notArray));
produces:
Exception in thread "main" java.lang.IllegalArgumentException: Argument is not an array
If you need to fail fast before invoking Array#getLength you can check if the argument is actually array:
if (!object.getClass().isArray()) {
// object is not array. Do something with it
}
answered Jan 26 at 18:40
caco3caco3
1,9181720
Thank you so much! So just for clarification: a primitive Array does not inherit fromObject, but a primitive value does inherit fromObject?
– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. Anintis not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.
– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is anObject, whether its elements areObjects or not.
– Boann
Jan 27 at 17:13
add a comment |
The caco3 answer is fine.
Note that if the Object params are not arrays you will discover it only at runtime.
A safer way would be to overload the method for primitives :
void validateSameSize(Object first, Object second) throws Exception {
if (first.length != second.length) throw new Exception();
}
void validateSameSize(int first, int second) throws Exception {
if (first.length != second.length) throw new Exception();
}
And so for...
It will require you to write more code but your code would be more robust.
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use Array#getLength:
public static boolean sameSize(Object arrayA, Object arrayB) {
return Array.getLength(arrayA) == Array.getLength(arrayB);
}
It will work with non-primitive arrays as well as with primitive ones:
System.out.println(sameSize(new int[0], new int[100])); // false
System.out.println(sameSize(new char[0], new int[0])); // true
System.out.println(sameSize(new Object[0], new Object[0])); // true
System.out.println(sameSize(new Object[0], new List[0])); // true
Also don't forget that passing an Object that is not an array to Array#getLength will result in IllegalArgumentException.
This code:
Object notArray = 100;
System.out.println(Array.getLength(notArray));
produces:
Exception in thread "main" java.lang.IllegalArgumentException: Argument is not an array
If you need to fail fast before invoking Array#getLength you can check if the argument is actually array:
if (!object.getClass().isArray()) {
// object is not array. Do something with it
}
answered Jan 26 at 18:40
caco3caco3
1,9181720
Thank you so much! So just for clarification: a primitive Array does not inherit fromObject, but a primitive value does inherit fromObject?
– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. Anintis not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.
– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is anObject, whether its elements areObjects or not.
– Boann
Jan 27 at 17:13
add a comment |
You can use Array#getLength:
public static boolean sameSize(Object arrayA, Object arrayB) {
return Array.getLength(arrayA) == Array.getLength(arrayB);
}
It will work with non-primitive arrays as well as with primitive ones:
System.out.println(sameSize(new int[0], new int[100])); // false
System.out.println(sameSize(new char[0], new int[0])); // true
System.out.println(sameSize(new Object[0], new Object[0])); // true
System.out.println(sameSize(new Object[0], new List[0])); // true
Also don't forget that passing an Object that is not an array to Array#getLength will result in IllegalArgumentException.
This code:
Object notArray = 100;
System.out.println(Array.getLength(notArray));
produces:
Exception in thread "main" java.lang.IllegalArgumentException: Argument is not an array
If you need to fail fast before invoking Array#getLength you can check if the argument is actually array:
if (!object.getClass().isArray()) {
// object is not array. Do something with it
}
answered Jan 26 at 18:40
caco3caco3
1,9181720
Thank you so much! So just for clarification: a primitive Array does not inherit fromObject, but a primitive value does inherit fromObject?
– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. Anintis not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.
– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is anObject, whether its elements areObjects or not.
– Boann
Jan 27 at 17:13
add a comment |
You can use Array#getLength:
public static boolean sameSize(Object arrayA, Object arrayB) {
return Array.getLength(arrayA) == Array.getLength(arrayB);
}
It will work with non-primitive arrays as well as with primitive ones:
System.out.println(sameSize(new int[0], new int[100])); // false
System.out.println(sameSize(new char[0], new int[0])); // true
System.out.println(sameSize(new Object[0], new Object[0])); // true
System.out.println(sameSize(new Object[0], new List[0])); // true
Also don't forget that passing an Object that is not an array to Array#getLength will result in IllegalArgumentException.
This code:
Object notArray = 100;
System.out.println(Array.getLength(notArray));
produces:
Exception in thread "main" java.lang.IllegalArgumentException: Argument is not an array
If you need to fail fast before invoking Array#getLength you can check if the argument is actually array:
if (!object.getClass().isArray()) {
// object is not array. Do something with it
}
answered Jan 26 at 18:40
caco3caco3
1,9181720
You can use Array#getLength:
public static boolean sameSize(Object arrayA, Object arrayB) {
return Array.getLength(arrayA) == Array.getLength(arrayB);
}
It will work with non-primitive arrays as well as with primitive ones:
System.out.println(sameSize(new int[0], new int[100])); // false
System.out.println(sameSize(new char[0], new int[0])); // true
System.out.println(sameSize(new Object[0], new Object[0])); // true
System.out.println(sameSize(new Object[0], new List[0])); // true
Also don't forget that passing an Object that is not an array to Array#getLength will result in IllegalArgumentException.
This code:
Object notArray = 100;
System.out.println(Array.getLength(notArray));
produces:
Exception in thread "main" java.lang.IllegalArgumentException: Argument is not an array
If you need to fail fast before invoking Array#getLength you can check if the argument is actually array:
if (!object.getClass().isArray()) {
// object is not array. Do something with it
}
answered Jan 26 at 18:40
caco3caco3
1,9181720
edited Jan 26 at 19:10
answered Jan 26 at 18:40
caco3caco3
1,9181720
answered Jan 26 at 18:40
caco3caco3
1,9181720
answered Jan 26 at 18:40
caco3caco3
1,9181720
1,9181720
Thank you so much! So just for clarification: a primitive Array does not inherit fromObject, but a primitive value does inherit fromObject?
– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. Anintis not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.
– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is anObject, whether its elements areObjects or not.
– Boann
Jan 27 at 17:13
add a comment |
Thank you so much! So just for clarification: a primitive Array does not inherit fromObject, but a primitive value does inherit fromObject?
– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. Anintis not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.
– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is anObject, whether its elements areObjects or not.
– Boann
Jan 27 at 17:13
Thank you so much! So just for clarification: a primitive Array does not inherit from
Object, but a primitive value does inherit from Object?– Schottky
Jan 26 at 18:45
Thank you so much! So just for clarification: a primitive Array does not inherit from
Object, but a primitive value does inherit from Object?– Schottky
Jan 26 at 18:45
No, primitive types do not take part in inheritance at all. An
int is not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.– Roland Illig
Jan 26 at 18:51
No, primitive types do not take part in inheritance at all. An
int is not an object, it is just the number. The Java Language Specification probably has a chapter called Primitive Types, which would be the definitive source on this topic.– Roland Illig
Jan 26 at 18:51
@Schottky Every Java array is an
Object, whether its elements are Objects or not.– Boann
Jan 27 at 17:13
@Schottky Every Java array is an
Object, whether its elements are Objects or not.– Boann
Jan 27 at 17:13
add a comment |
The caco3 answer is fine.
Note that if the Object params are not arrays you will discover it only at runtime.
A safer way would be to overload the method for primitives :
void validateSameSize(Object first, Object second) throws Exception {
if (first.length != second.length) throw new Exception();
}
void validateSameSize(int first, int second) throws Exception {
if (first.length != second.length) throw new Exception();
}
And so for...
It will require you to write more code but your code would be more robust.
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
add a comment |
The caco3 answer is fine.
Note that if the Object params are not arrays you will discover it only at runtime.
A safer way would be to overload the method for primitives :
void validateSameSize(Object first, Object second) throws Exception {
if (first.length != second.length) throw new Exception();
}
void validateSameSize(int first, int second) throws Exception {
if (first.length != second.length) throw new Exception();
}
And so for...
It will require you to write more code but your code would be more robust.
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
add a comment |
The caco3 answer is fine.
Note that if the Object params are not arrays you will discover it only at runtime.
A safer way would be to overload the method for primitives :
void validateSameSize(Object first, Object second) throws Exception {
if (first.length != second.length) throw new Exception();
}
void validateSameSize(int first, int second) throws Exception {
if (first.length != second.length) throw new Exception();
}
And so for...
It will require you to write more code but your code would be more robust.
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
The caco3 answer is fine.
Note that if the Object params are not arrays you will discover it only at runtime.
A safer way would be to overload the method for primitives :
void validateSameSize(Object first, Object second) throws Exception {
if (first.length != second.length) throw new Exception();
}
void validateSameSize(int first, int second) throws Exception {
if (first.length != second.length) throw new Exception();
}
And so for...
It will require you to write more code but your code would be more robust.
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
answered Jan 26 at 19:01
davidxxxdavidxxx
68.7k67399
68.7k67399
add a comment |
add a comment |
