In normal maths terms, -1756046391 * -1291488517 will equal 2267913749295792147.
When i enter the exact same equation in java, i get the answer: -1756046391 * -1291488517 = 19.
Can anyone shed any light on this?
Overflow.
Check out Integer.MAX_VALUE. An Integer in Java is a 32 bit 2s complement value. You can't exceed the MAX_VALUE.
When dealing with very large numbers, you need to make sure the data type you used is big enough to store that number. In java you have these primitive number types:
type: min: max:
byte -128 127
short -32768 32767
int -2,147,483,648 2,147,483,647
long -9,223,372,036,854,775,808 9,223,372,036,854,775,807
So as you can see, your number would just about fit into a long. But you're bound to go over that, so you should probably use the BigInt class instead:
http://docs.oracle.com/javase/6/docs/api/java/math/BigInteger.html
Either use longs as others said, or if that is not enough for your application, consider using BigInteger.
Post the exact code you used to reproduce this.
You are most likely using ints. Use longs, because 2267913749295792147 is too big to fit in an int.
Related
What is an integer overflow error?
Why do i care about such an error?
What are some methods of avoiding or preventing it?
Integer overflow occurs when you try to express a number that is larger than the largest number the integer type can handle.
If you try to express the number 300 in one byte, you have an integer overflow (maximum is 255). 100,000 in two bytes is also an integer overflow (65,535 is the maximum).
You need to care about it because mathematical operations won't behave as you expect. A + B doesn't actually equal the sum of A and B if you have an integer overflow.
You avoid it by not creating the condition in the first place (usually either by choosing your integer type to be large enough that you won't overflow, or by limiting user input so that an overflow doesn't occur).
The easiest way to explain it is with a trivial example. Imagine we have a 4 bit unsigned integer. 0 would be 0000 and 1111 would be 15. So if you increment 15 instead of getting 16 you'll circle back around to 0000 as 16 is actually 10000 and we can not represent that with less than 5 bits. Ergo overflow...
In practice the numbers are much bigger and it circles to a large negative number on overflow if the int is signed but the above is basically what happens.
Another way of looking at it is to consider it as largely the same thing that happens when the odometer in your car rolls over to zero again after hitting 999999 km/mi.
When you store an integer in memory, the computer stores it as a series of bytes. These can be represented as a series of ones and zeros.
For example, zero will be represented as 00000000 (8 bit integers), and often, 127 will be represented as 01111111. If you add one to 127, this would "flip" the bits, and swap it to 10000000, but in a standard two's compliment representation, this is actually used to represent -128. This "overflows" the value.
With unsigned numbers, the same thing happens: 255 (11111111) plus 1 would become 100000000, but since there are only 8 "bits", this ends up as 00000000, which is 0.
You can avoid this by doing proper range checking for your correct integer size, or using a language that does proper exception handling for you.
An integer overflow error occurs when an operation makes an integer value greater than its maximum.
For example, if the maximum value you can have is 100000, and your current value is 99999, then adding 2 will make it 'overflow'.
You should care about integer overflows because data can be changed or lost inadvertantly, and can avoid them with either a larger integer type (see long int in most languages) or with a scheme that converts long strings of digits to very large integers.
Overflow is when the result of an arithmetic operation doesn't fit in the data type of the operation. You can have overflow with a byte-sized unsigned integer if you add 255 + 1, because the result (256) does not fit in the 8 bits of a byte.
You can have overflow with a floating point number if the result of a floating point operation is too large to represent in the floating point data type's exponent or mantissa.
You can also have underflow with floating point types when the result of a floating point operation is too small to represent in the given floating point data type. For example, if the floating point data type can handle exponents in the range of -100 to +100, and you square a value with an exponent of -80, the result will have an exponent around -160, which won't fit in the given floating point data type.
You need to be concerned about overflows and underflows in your code because it can be a silent killer: your code produces incorrect results but might not signal an error.
Whether you can safely ignore overflows depends a great deal on the nature of your program - rendering screen pixels from 3D data has a much greater tolerance for numerical errors than say, financial calculations.
Overflow checking is often turned off in default compiler settings. Why? Because the additional code to check for overflow after every operation takes time and space, which can degrade the runtime performance of your code.
Do yourself a favor and at least develop and test your code with overflow checking turned on.
From wikipedia:
In computer programming, an integer
overflow occurs when an arithmetic
operation attempts to create a numeric
value that is larger than can be
represented within the available
storage space. For instance, adding 1 to the largest value that can be represented
constitutes an integer overflow. The
most common result in these cases is
for the least significant
representable bits of the result to be
stored (the result is said to wrap).
You should care about it especially when choosing the appropriate data types for your program or you might get very subtle bugs.
From http://www.first.org/conference/2006/papers/seacord-robert-slides.pdf :
An integer overflow occurs when an integer is
increased beyond its maximum value or
decreased beyond its minimum value.
Overflows can be signed or unsigned.
P.S.: The PDF has detailed explanation on overflows and other integer error conditions, and also how to tackle/avoid them.
I'd like to be a bit contrarian to all the other answers so far, which somehow accept crappy broken math as a given. The question is tagged language-agnostic and in a vast number of languages, integers simply never overflow, so here's my kind-of sarcastic answer:
What is an integer overflow error?
An obsolete artifact from the dark ages of computing.
why do i care about it?
You don't.
how can it be avoided?
Use a modern programming language in which integers don't overflow. (Lisp, Scheme, Smalltalk, Self, Ruby, Newspeak, Ioke, Haskell, take your pick ...)
I find showing the Two’s Complement representation on a disc very helpful.
Here is a representation for 4-bit integers. The maximum value is 2^3-1 = 7.
For 32 bit integers, we will see the maximum value is 2^31-1.
When we add 1 to 2^31-1 : Clockwise we move by one and it is clearly -2^31 which is called integer overflow
Ref : https://courses.cs.washington.edu/courses/cse351/17wi/sections/03/CSE351-S03-2cfp_17wi.pdf
This happens when you attempt to use an integer for a value that is higher than the internal structure of the integer can support due to the number of bytes used. For example, if the maximum integer size is 2,147,483,647 and you attempt to store 3,000,000,000 you will get an integer overflow error.
I am a little confused on when to use what primitives. If I am defining a number, how do I know what to use byte, short, int, or long? I know they are different bytes, but does that mean that I can only use one of them for a certain number?
So simply, my question is, when do I use each of the four primitives listed above?
If I am lets say defining a number, how do I know what to use byte, short, int, or long?
Depending on your use-case. There's no huge penalty to using an int as opposed to a short, unless you have billions of numbers. Simply consider the range a variable might use. In most cases it is reasonable to use int, whose range is -2,147,483,648 to 2,147,483,647, while long handles numbers in the range of +/- 9.22337204*1018. If you aren't sure, long won't specifically hurt.
The only reasons you might want to use byte specifically is if you are storing byte data such as parts of a file, or are doing something like network communication or serialization where the number of bytes is important. Remember that Java's bytes are also signed (-128 to 127). Same for short--might be useful to save 2GB of memory for a billion-element array, but not specifically useful for much other than, again, serialization with a specific byte alignment.
does that mean that I can only use one of them for a certain number?
No, you can use any that is large enough to handle that number. Of course, decimal values need a double or a float--double is usually ideal due to its higher precision and few drawbacks. Some libraries (e.g. 3D drawing) might use floats however. Remember that you can cast numeric types (e.g. (byte) someInt or (float) functionReturningADouble();
A byte can be a number from -128 to 127
A short can be a number from -32,768 to 32,767
An int can be from -2^32 to 2^32 - 1
A long is -2^63 to 2^63 - 1
Usually unless space is an issue an int is fine.
In order to use a primitive you need to know the data range of the number stored in it. This depends heavily on the thing that you are modeling:
Use byte when your number is in the range [-128..127]
Use short when your number is in the range [-32768..32767]
Use int when your number is in the range [-231..231-1]
Use long when your number is in the range [-263..263-1]
In addition, byte data type is used for representing "raw" binary data, in which case it is not used like a number.
When you want an unlimited range, use BigInteger. This flexibility comes at a cost: operations on BigInteger can be orders of magnitude more expensive than identical operations on primitives.
Do you know what values your variable will potentially hold?
If it will be between -128 and 127, use a byte.
If it will be between -32,768 and 32,767, use a short.
If it will be between -2^32 and 2^32 - 1, use an int.
If it will be between -2^63 and 2^63 - 1, use a long.
Generally an int will be fine for most cases, but you might use a byte or a short if memory is a concern, or a long if you need larger values.
Try writing a little test program. What happens when you put a value that's too large in one of these types of variables?
Source: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
See also: the year 2038 problem
The general guidance should definitely be: If you don't have any good reason to pick a specific primitive, pick int.
This has at least three advantages:
Every kind of arithmetic or logical operation will return an int. Hence if you use anything smaller you will have to cast a lot to actually assign values back.
It's the accepted standard for "some number". If you pick anything else, other developers will wonder about the reasons.
It will be the most efficient, although it's likely that the performance difference from using anything else will be minuscule.
That said if you have an array of millions of elements yes you definitely should figure out what primitive fits the input range - I refer you to the other answers for a summary of how large each one is.
#dasblinkenlight
When you are working for large project, you need to optimize the code for each requirement. Apparently you would reduce the memory occupied by the project.
Hence using int, short, long, byte plays a vital role in terms of optimising the memory. When the interger holds a 1 digit value then you declare the integer with int or short, but you can still choose long,float and double technically but to become a good programmer you can try to follow standards.
Byte - Its 8 bit. So it can hold a value from -128 to 127.
short: Its 32 bit. so -32,768 and a maximum value of 32,767
Long- Its 64 bit. -2 (power of) 63 and a maximum value of 2(power of)63-1
float and double- for large decimal values.
Boolean for Yes or No decesions- can hold a value of 1 or 0
Char for characters.
Hope this answer helps you.
I'm porting some C++ code to Java code.
There is no unsigned datatype in java which can hold 64 bits.
I have a hashcode which is stored in Java's long datatype (which of course is signed).
long vp = hashcode / 38; // hashcode is of type 'long'
Since 38 here is greater than 2, the resulting number can be safely used for any other arithmetic in java.
The question is what if the signed bit in 'hashcode' is set to 1. I don't want to get a negative value in variable vp. I wanted a positive value as if the datatype is an unsigned one.
P.S: I don't want to used Biginteger for this purpose because of performance issues.
Java's primative integral types are considered signed, and there isn't really anything you can do about it. However, depending on what you need it for, this may not matter.
Since the integers are all done in two's complement, signed and unsigned are exact same at the binary level. The difference is how you interpret them, and in certain operations. Specifically, right shift, division, modulus and comparison differ. Unsigned right shifts can be done with the >>> operator. As long as you don't need one of the missing operators, you can use longs perfectly well.
If you can use third-party libraries, you can e.g. use Guava's UnsignedLongs class to treat long values as unsigned for many purposes, including division. (Disclosure: I contribute to Guava.)
Well here is how i solved this. Right shift hashcode by 1 bit(division by 2). Then divide that right shifted number by 19(which is 38/2). So essentially i divided the number by 38 exactly like how it is done in c++. I got the same value what i got in c++
I know this is a n00b question but I want to define a variable with a value between 0 and 100 in JAVA. I can use INT or BYTE as data types - but which one would be the best to use. And why?(benefits?)
Either int or byte would work for storing a number in that range.
Which is best depends on what you are aiming to do.
If you need to store lots of these numbers in an array, then a byte[] will take less space than an int[] with the same length.
(Surprisingly) a byte variable takes the same amount of space as a int ... due to the way that the JVM does the stack frame / object frame layout.
If you are doing lots of arithmetic, etc using these values, then int is more convenient than byte. In Java, arithmetic and bitwise operations automatically promote the operands to int, so when you need to assign back to a byte you need to use a (byte) type-cast.
Use byte datatype to store value between -128 to 127
I think since you have used 0.00 with decimals, this means you want to allow decimals.
The best way to do this is by using a floating point type: double or float.
However, if you don't need decimals, and the number is always an integer, you can use either int or byte. Byte can hold anything up to +127 or down to -128, and takes up much less space than an int.
with 32-bit ints on your system, your byte would take up 1/4 of the space!
The disadvantages are that you have to be careful when doing arithmetic on bytes - they can overflow, and cause unexpected results if they go out of range.
If you are using basic IO functions, like creating sounds or reading old file formats, byte is invaluable. But watch out for the sign convention: negative has the high-order bit set. However when doing normal numerical calculations, I always use int, since it allows easy extension to larger values if needed, and 32-bit cpus there is minimal speed cost really. Only thing is if you are storing a large quantity of them.
Citing Primitive Data Types:
byte: The byte data type is an 8-bit signed two's complement integer. It has a minimum value of -128 and a maximum value of 127 (inclusive). The byte data type can be useful for saving memory in large arrays, where the memory savings actually matters. They can also be used in place of int where their limits help to clarify your code; the fact that a variable's range is limited can serve as a form of documentation.
int: The int data type is a 32-bit signed two's complement integer. It has a minimum value of -2,147,483,648 and a maximum value of 2,147,483,647 (inclusive). For integral values, this data type is generally the default choice unless there is a reason (like the above) to choose something else. This data type will most likely be large enough for the numbers your program will use, but if you need a wider range of values, use long instead.
Well it depends on what you're doing... out of the blue like that I would say use int because it's a standard way to declare an integer. But, if you do something specific that require memory optimization than maybe byte could be better in that case.
Technically, neither one is appropriate--if you need to store the decimal part, you want a float.
I'd personally argue that unless you're actually getting into resource-constrained issues (embedded system, big data, etc.) your code will be slightly clearer if you go with int. Good style is about being legible to people, so indicating that you're working with integers might make more sense than prematurely optimizing and thus forcing future maintainers of your code (including yourself) to speculate as to the true nature of that variable.
Hey , i hope i get help with this.
im a coder of a rsps (runescape private server)
and in this game you could like have items and weapons
and the max anmount of a item you can have is 2147000000
and i can change the amount of the max by changing this int
public int maxItemAmount = 2147000000;
and it works
but i want to make it like 3000000000
and i do this
public int maxItemAmount = 3000000000;
and when i compile i get this error
integrer number too large: 3000000000
please guys help me out if you can :)
Integer has an upper bound of 2^31 (2147483648). If you want numbers longer than that, you can use a long or double.
Integers are signed 32-bit values and thus can have a maximum value of 231 = (note that one bit is used for the sign).
You need to change the type of maxItemAmount to long.
you should go for 64 bit types ... i.e: long ...
do ...
public long maxItemAmount = 3000000000L;
A 32-bit signed integer has a range of -232 to 232-1, or − 2,147,483,648 to 2,147,483,647. If you want an integral value outside this range you need to use a 64-bit (long) variable. On the other hand, how likely is it that anyone will have over 2 billion distinct items or weapons? Perhaps, you want to rethink it and keep track of items and their quantities separately. You might also want to consider that changing to use a long may have unexpected consequences if parts of the code assume that it's a 32-bit value.
You can check out the max value of numbers with Integer.MAX_VALUE and Long.MAX_VALUE.
If a long is still not enough for your needs, you can check out the BigInteger class.
Instead of int you should use long. Here you can find more information about the precision of Java primitive data types.
Quoting from the link:
long: The long data type is a 64-bit signed two's complement integer. It has a minimum value of -9,223,372,036,854,775,808 and a maximum value of 9,223,372,036,854,775,807 (inclusive). Use this data type when you need a range of values wider than those provided by int.
As other people has sayed, you have to use another data type that support big numbers.
I recommend you using long. For example:
public long maxItemAmount = 3000000000L;
Notice the L at the end of the value. It tells the runtime that 3000000000 is a long value. Or use:
public long maxItemAmount = new Long("3000000000");
See more info on this page.
Also, someone commented using a BigInteger. It's like a String; don't have limit.
However, I recommend you to use it rarely. A long will probably suit your needs.
EDIT: Strictly speaking, BigIntegers and Strings have limit (see comments).