This question already has answers here:
Do Java arrays have a maximum size?
(10 answers)
What does "possible lossy conversion" mean and how do I fix it?
(1 answer)
Why I can't create an array with large size?
(5 answers)
Closed 1 year ago.
I wish to enter one int and another long ex: 1 and 1000000000, and now I wish to create an array of size 1000000000. And then at each index of array, store int val, ex: arr[100000000] = 4.
When I'm trying to do this Netbeans shows me an error in this line:
arr = new long[y+1]` and `arr[j] = 0`
"Possible Lossy conversion from long to int".
Here's my code:-
public static void main(String[] args) throws IOException
{
BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
String[] xe = s.readLine().split(" ");
int x = Integer.parseInt(xe[0]);
long y = Long.parseLong(xe[1]);
long []arr;
arr = new long[y+1];
for(long j=0;j<=y;j++)
arr[j] = 4;
}
Array index is an integer in Java and the compiler will advice you. So maximum array size is (aproximately) Integer.MAX_VALUE. For bigger arrays you should use ArrayList.
To go around this dirt&quick
arr = new long[(int)y+1];
The size of an array can only be an int. That is you, can not create arrays that are larger than Integer.MAX_VALUE (2147483647), probably a few elements less (depending on the VM). You can cast your value to int
arr = new long[(int)(y+1)];
but this will produce invalid results when the value of y is actually larger than the maximum allowed size.
You need to convert/cast y and j to int.
Also your incrementing var shouldn't be long when you're just adding 1.
You cannot create an array with more than 2^31-1 entries, so you should use an int value when you do so (the compiler is simply warning you that the size will be truncated from long to int). 1000000000 is small enough to fit into int, so you basically have no reason to use long y in order to store this value in the first place.
According to your description, the array itself is designated to store int values, so it doesn't need to be an array of long values.
In short, you can and should change every long in your code to int.
I think you have some misconception about typecasting here. In Java down casting is allowed as already mentioned you can down cast it to integer
arr = new long[(int)(y+1)];
But here the main problem is that array allocation should takes integer value so that the same number of homogeneous space can be allocated.
If you want more space than an integer range then you should use ArrayList which is dynamically grow-able array because if we give that much liability to a user to declare a large amount array memory then our system may run out of the memory as array is a static memory allocation data structure.
The same stands true for
arr[j]=4;
It should be:
arr[(int)j]=4;
In static array we cannot use long datatype values for assigning index position.
1.Either convert long to int in for/while/do while loop.
2. Typecast arr[(int)i] to remove error
Related
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Initialize a long in Java
(4 answers)
The literal xyz of type int is out of range
(5 answers)
Why, In Java arithmetic, overflow or underflow will never throw an Exception?
(4 answers)
Closed 4 years ago.
Here are my 2 examples:
1. Directly assigned value to long data type:
long a = 12243221112432;
I get an error:
integer number too large
But when assisgned like this:
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(12);
al.add(24);
al.add(32);
al.add(21);
al.add(11);
al.add(24);
al.add(32);
long a=0;
for(int i=0; i<al.size(); i++){
a = a*100 + al.get(i);
}
System.out.println(a);
I get this output:
12243221112432
Why doesn't java throw an error in second example?
It isn't allowing to assign large value directly(example 1) but indirectly(example 2) it stores it and also allows me to use it too!
What is the reason for this to occur?
Is it because i am using integer in arraylist or something else?
UPDATE
Why is the large value stored in 'long a' in second example without using literal L?
It should have given me an error during 5th or 6th iteration of for loop...
Note
My question is not regarding the first example... I am asking why it worked for the second example...
Dont mark the question duplicate, since the other questions do not have my answer..stated above
For your assignment to work, you need to write it that way
long a = 12243221112432L;
This will indicate the compiler that your input is a long, not an int
integer number too large
as the error said, the number 12243221112432 is too large for an integer, it does not says that this number cannot fit into a long
Te be able to have this number, you have to make it a long by using the l or L indice : 12243221112432L
long : l L : 1716L
float : f F : 3.69F
double : d D : 6936D
I am new to Java programming and I am working with arrays .In arrays
index needs to be an integer and it doesn't allow float or double so I used long data type for index and it gave error. Whereas when I used byte and short and Int it worked . I want to know since the error was "possible lossy conversion from long to int "
Is it such that only int is allowed in index and since byte and short are small in size that's why it worked (auto promotion) and long was larger than int so it gave error(no auto depromotion)plz help
import Java.util.*;
class Demo{
public static void main(String args[]){
long n=5;
int a[]=new int[n]; //error possible lossy conversion from long to int
System.out.println(Arrays.toString(a));
}
}
In short, it's because that's the way the language is designed. If you take a look at section 10.7 of the Java language specifications, you'll notice that the length of the array is defined as an int (which is smaller than a long). Semtantically, ints and longs represent integers, but "larger" integer types will not be automatically cast to "smaller" integer types.
Needs to be int. Long, short, byte, float are other data types. Implicit or explicit conversion of your supplied data type can occur, but the index of your array will only be constructed with an int by the javac.
An array object contains a number of variables. The number of variables may be zero, in which case the array is said to be empty. The variables contained in an array have no names; instead they are referenced by array access expressions that use non-negative integer index values. These variables are called the components of the array. If an array has n components, we say n is the length of the array; the components of the array are referenced using integer indices from 0 to n - 1, inclusive.
Oracle Docs
You cannot cast long to int or rather cannot convert from long to int.
The only conversion you can do is
•
byte to short, int, long, float, or double
•
short to int, long, float, or double
•
char to int, long, float, or double
•
int to long, float, or double
•
long to float or double
•
float to double
Since you have array of int you can store int ,short and byte datatype values
This question already has answers here:
How does Java handle integer underflows and overflows and how would you check for it?
(12 answers)
Closed 7 years ago.
I wrote a program to store Fibonacci numbers and will retrieve the nth Fibonacci number. It works fine until the 50th Fibonacci where it returns a negative number.
getFibonacci(47) returns 1836311903 but
getFibonacci(48) returns -1323752223. Why is this?
public class Fibonacci {
static HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
public static void main(String[] args) {
int x;
storeFibonacciNumbers(50);
System.out.println(getFibonacci(48));
}
public static void storeFibonacciNumbers (int n){
for (int x=1; x <= n; x++ ){
if(x == 1){
map.put(x, 0);
}
else if (x == 2) {
map.put(x, x-1);
}
else
map.put(x, map.get(x-1) + map.get(x-2));
}
}
public static long getFibonacci (int x){
return map.get(x);
}
}
This is because the maximum value of an Integer has been reached. When you try to add a value to it, it overflows, and "wraps around" to the negative.
An alternative is to use a type that supports higher maximums (e.g. Long).
At some point, you might be forced to switch to types that are more complex, but are built specifically to store very large integers (e.g. BigInteger).
In the HashMap you are storing Fibonacci number as a Integer. The maximum value that an Int can hold is 2147483647 (2^31-1)
So based on your calculations Fibonacci(47) must exceed that maximum value and overflow is happening.
Change your HashMap value type to Long. It will fix your problem.
An Integer Overflow is the condition that occurs when the result of an arithmetic operation, such as multiplication or addition, exceeds the maximum size of the integer type used to store it.
It's because the int type only goes so high. The max value is 2147483647, at which point it will just start over. You need to use a different datatype, such as a long if you want to represent a bigger number correctly.
The max java integer is 2^31-1 or 2147483647. After you reach that value you will start returning negative numbers.
You can use the constant Integer.MAX_VALUE as a way to stop your program once you reach that value.
This is called Integer Overflow, and occurs when a number increments over/falls under its maximal/minimal (memory) bounds (for java.lang.Integer this would be -2^31 and 2^31-1)
I was testing how Java (SE7) would deal with the int that exceed its maximum value through the following code:
int index = 2147483647;//the maximum value of int
long size = 2147483648L; //More than the maximum value of int by 1
int safeCounter=0; //To prevent the infinite loop
while (index<size)
{
System.out.println("Index now is : "+index);//show the int value
index++; //increment the int value
safeCounter++; //increment the number of desired loops
if (safeCounter==3){
break;//to break the loop after 3 turns
}
}
and what I got is:
Index now is : 2147483647
Index now is : -2147483648
Index now is : -2147483647
So after being confused by this, (which if I don't use the safeCounter it would keep going forever between the maximum value and the minimum value of int -- and no exception is thrown) I was wondering how would an ArrayList handle a situation where the the number of elements exceed the maximum value of int (assuming that the heap space is not an issue)?
And if ArrayList can't handle this, Is there other data structure which can?
Can you also explain the behavior I got from the int variable?
Can an ArrayList contain more elements than the maximum value of int?
In practice no. An ArrayList is backed by a single Java array, and the maximum size of an array is Integer.MAX_VALUE.
(Hypothetically, Oracle could redo the implementation of ArrayList to use an array of arrays without breaking user code. But the chances of them doing that are pretty small.)
A LinkedList can handle as many elements as you can represent in memory. Or you could implement your own list type. Indeed, you could even implement a list type that can hold more elements than you could store in memory ... or even an unbounded number of elements if your list is actually a generator.
The fact that size() returns an int result (etcetera) is not actually an impediment. The List API spec deals with this anomaly.
The behaviour of your code is simply explained. Integer arithmetic in Java has silent overflow. If you add 1 to the largest positive value for an integer type, it wraps around to the largest negative value; i.e. MAX_VALUE + 1 == MIN_VALUE ... for integer types.
ArrayList can't handle that.Maximum limit of arraylist size is Integer.MAX_VALUE.You can use LinkedList which can contain any number of elements(depends on your memory actually):-)
From ArrayList.java:
**
* The array buffer into which the elements of the ArrayList are stored.
* The capacity of the ArrayList is the length of this array buffer.
*/
private transient Object[] elementData;
As it uses array in its implementation, you cannot index beyond Integer.MAX_VALUE, so that's a limit.
For the int behviour, you can take a look at this question.
This is because Java uses signed integers. ArrayList index starts from 0 and there is no way to provide a negative index to the ArrayList.
One possible solution to your problem is, first convert the unsigned integer to signed and the n use it in ArrayList.
You could convert the signed to unsigned by using the following snippet:
public static long getUnsigned(int signed) {
if(signed > 0) return signed;
long signedVal = (long)(Math.pow(2, 32)) + signed;
return signedVal;
}
This question already has answers here:
How do I convert a String to an int in Java?
(47 answers)
Closed 5 years ago.
I have written a function to convert string to integer
if ( data != null )
{
int theValue = Integer.parseInt( data.trim(), 16 );
return theValue;
}
else
return null;
I have a string which is 6042076399 and it gave me errors:
Exception in thread "main" java.lang.NumberFormatException: For input string: "6042076399"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:461)
Is this not the correct way to convert string to integer?
Here's the way I prefer to do it:
Edit (08/04/2015):
As noted in the comment below, this is actually better done like this:
String numStr = "123";
int num = Integer.parseInt(numStr);
An Integer can't hold that value. 6042076399 (413424640921 in decimal) is greater than 2147483647, the maximum an integer can hold.
Try using Long.parseLong.
That's the correct method, but your value is larger than the maximum size of an int.
The maximum size an int can hold is 231 - 1, or 2,147,483,647. Your value is 6,042,076,399. You should look at storing it as a long if you want a primitive type. The maximum value of a long is significantly larger - 263 - 1. Another option might be BigInteger.
That string is greater than Integer.MAX_VALUE. You can't parse something that is out of range of integers. (they go up to 2^31-1, I believe).
In addition to what the others answered, if you have a string of more than 8 hexadecimal digits (but up to 16 hexadecimal digits), you could convert it to a long using Long.parseLong() instead of to an int using Integer.parseInt().