In C, the integer (for 32 bit machine) is 32 bits, and it ranges from -32,768 to +32,767.
In Java, the integer(long) is also 32 bits, but ranges from -2,147,483,648 to +2,147,483,647.
I do not understand how the range is different in Java, even though the number of bits is the same. Can someone explain this?
In C, the language itself does not determine the representation of certain datatypes. It can vary from machine to machine, on embedded systems the int can be 16 bit wide, though usually it is 32 bit.
The only requirement is that short int <= int <= long int by size. Also, there is a recommendation that int should represent the native capacity of the processor.
All types are signed. The unsigned modifier allows you to use the highest bit as part of the value (otherwise it is reserved for the sign bit).
Here's a short table of the possible values for the possible data types:
width minimum maximum
signed 8 bit -128 +127
signed 16 bit -32 768 +32 767
signed 32 bit -2 147 483 648 +2 147 483 647
signed 64 bit -9 223 372 036 854 775 808 +9 223 372 036 854 775 807
unsigned 8 bit 0 +255
unsigned 16 bit 0 +65 535
unsigned 32 bit 0 +4 294 967 295
unsigned 64 bit 0 +18 446 744 073 709 551 615
In Java, the Java Language Specification determines the representation of the data types.
The order is: byte 8 bits, short 16 bits, int 32 bits, long 64 bits. All of these types are signed, there are no unsigned versions. However, bit manipulations treat the numbers as they were unsigned (that is, handling all bits correctly).
The character data type char is 16 bits wide, unsigned, and holds characters using UTF-16 encoding (however, it is possible to assign a char an arbitrary unsigned 16 bit integer that represents an invalid character codepoint)
width minimum maximum
SIGNED
byte: 8 bit -128 +127
short: 16 bit -32 768 +32 767
int: 32 bit -2 147 483 648 +2 147 483 647
long: 64 bit -9 223 372 036 854 775 808 +9 223 372 036 854 775 807
UNSIGNED
char 16 bit 0 +65 535
In C, the integer(for 32 bit machine) is 32 bit and it ranges from -32768 to +32767.
Wrong. 32-bit signed integer in 2's complement representation has the range -231 to 231-1 which is equal to -2,147,483,648 to 2,147,483,647.
A 32 bit integer ranges from -2,147,483,648 to 2,147,483,647. However the fact that you are on a 32-bit machine does not mean your C compiler uses 32-bit integers.
The C language definition specifies minimum ranges for various data types. For int, this minimum range is -32767 to 32767, meaning an int must be at least 16 bits wide. An implementation is free to provide a wider int type with a correspondingly wider range. For example, on the SLES 10 development server I work on, the range is -2147483647 to 2137483647.
There are still some systems out there that use 16-bit int types (All The World Is Not A VAX x86), but there are plenty that use 32-bit int types, and maybe a few that use 64-bit.
The C language was designed to run on different architectures. Java was designed to run in a virtual machine that hides those architectural differences.
The strict equivalent of the java int is long int in C.
Edit:
If int32_t is defined, then it is the equivalent in terms of precision. long int guarantee the precision of the java int, because it is guarantee to be at least 32 bits in size.
The poster has their java types mixed up.
in java, his C in is a short:
short (16 bit) = -32768 to 32767
int (32 bit) = -2,147,483,648 to 2,147,483,647
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
That's because in C - integer on 32 bit machine doesn't mean that 32 bits are used for storing it, it may be 16 bits as well. It depends on the machine (implementation-dependent).
Actually the size in bits of the int, short, long depends on the compiler implementation.
E.g. on my Ubuntu 64 bit I have short in 32 bits, when on another one 32bit Ubuntu version it is 16 bit.
It is actually really simple to understand, you can even compute it with the google calculator:
you have 32 bits for an int and computers are binary, therefore you can have 2 values per bit (spot).
if you compute 2^32 you will get the 4,294,967,296. so if you divide this number by 2, (because half of them are negative integers and the other half are positive), then you get 2,147,483,648. and this number is the biggest int that can be represented by 32 bits, although if you pay attention you will notice that 2,147,483,648 is greater than 2,147,483,647 by 1, this is because one of the numbers represents 0 which is right in the middle unfortunately 2^32 is not an odd number therefore you dont have only one number in the middle, so the possitive integers have one less cipher while the negatives get the complete half 2,147,483,648.
And thats it. It depends on the machine not on the language.
In C range for __int32 is –2147483648 to 2147483647. See here for full ranges.
unsigned short 0 to 65535
signed short –32768 to 32767
unsigned long 0 to 4294967295
signed long –2147483648 to 2147483647
There are no guarantees that an 'int' will be 32 bits, if you want to use variables of a specific size, particularly when writing code that involves bit manipulations, you should use the 'Standard Integer Types'.
In Java
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).
in standard C, you can use INT_MAX as the maximum 'int' value, this constant must be defined in "limits.h". Similar constants are defined for other types (http://www.acm.uiuc.edu/webmonkeys/book/c_guide/2.5.html), as stated, these constant are implementation-dependent but have a minimum value according to the minimum bits for each type, as specified in the standard.
Related
I'm confused about LEB128 or Little Endian Base 128 format. In the AOSP source code Leb128.java, its read function's return type whether signed or unsigned is int. I know the the size of int in java is 4 bytes aka 32bits. But the max length of LEB128 in AOSP is 5 bytes aka 35 bits. So where are the other lost 3bits.
Thanks for your reply.
Each byte of data in LEB only accounts for 7 bits in the actual output - the remaining bit is used to indicate whether or not it's the end.
From Wikipedia:
To encode an unsigned number using unsigned LEB128 first represent the number in binary. Then zero extend the number up to a multiple of 7 bits (such that the most significant 7 bits are not all 0). Break the number up into groups of 7 bits. Output one encoded byte for each 7 bit group, from least significant to most significant group.
The extra bits aren't so much "lost" as "used to indicate whether or not it's the end of the data".
You can't hope to encode arbitrary 32-bit values and some of them taking less than 4 bytes without some of them taking more than 4 bytes.
I can store numbers ranging from -127 to 127 but other than that it is impossible and the compiler give warning. The binary value of 127 is 01111111, and 130 is 10000010 still the same size (8 bits) and what I think is I can store 130 in Byte but it is not possible. How did that happen?
Java does not have unsigned types, each numeric type in Java is signed (except char but it is not meant for representing numbers but unicode characters).
Let's take a look at byte. It is one byte which is 8 bits. If it would be unsigned, yes, its range would be 0..255.
But if it is signed, it takes 1 bit of information to store the sign (2 possible values: + or -), which leaves us 7 bits to store the numeric (absolute) value. Range of 7 bit information is 0..127.
Note that the representation of signed integer numbers use the 2's complement number format in most languages, Java included.
Note: The range of Java's byte type is actually -128..127. The range -127..127 only contains 255 numbers (not 256 which is the number of all combinations of 8 bits).
In Java, a byte is a signed data type. You are thinking about unsigned bytes, in which case it is possible to store the value 130 in 8 bits. But with a signed data type, that also allows negative numbers, the first bit is necessary to indicate a negative number.
There are two ways to store negative numbers, (one's complement and two's complement) but the most popular one is two's complement. The benefit of it is that for two's complement, most arithmetic operations do not need to take the sign of the number into account; they can work regardless of the sign.
The first bit indicates the sign of the number: When the first bit is 1, then the number is negative. When the first bit is 0, then the number is positive. So you basically only have 7 bits available to store the magnitude of the number. (Using a small trick, this magnitude is shifted by 1 for negative numbers - otherwise, there would be two different bit patterns for "zero", namely 00000000 and 10000000).
When you want to store a number like 130, whose binary representation is 10000010, then it will be interpreted as a negative number, due to the first bit being 1.
Also see http://en.wikipedia.org/wiki/Two%27s_complement , where the trick of how the magnitude is shifted is explained in more detail.
I'm checking the Java doc, and seeing that Integer.MIN_VALUE is -2^31:
A constant holding the minimum value an int can have, -2^31.
While in C++, the 32 bits signed integer "long" has a different MIN value:
LONG_MIN: Minimum value for an object of type long int -2147483647
(-2^31+1) or less*
I am very confused why they are different and how Java get -2^31? In Java, if an integer has 32 bits and the first bit is used for sign, -2^31+1 is more logical, isn't it?
The value of the most significant bit in a 32 bit number is 2^31, and as this is negative in a signed integer the value is -2^31. I guess C++ uses -2^31+1 as the MIN_VALUE because this means it has the same absolute value as MAX_VALUE i.e. 2^31-1. This means an integer could store -MIN_VALUE, which is not the case in Java (which can cause some fun bugs).
C and C++ are intended to work with the native representation on machines using 1s or 2s complement or signed magnitude. With 2s complement you get a range from -2 n through 2 n -1, about like in Java. Otherwise you lose the most negative number and gain a representation for -0.
In C++ you're only guaranteed that n (the number of bits)is at least 16 where Java guarantees it's exactly 32.
I have an application in C and need to re-write it to Java. I have a problem with data types. I am not sure which data type is equal to the C ones in Java.
There are those I need to find out equal datatypes in java programming language for.
unsigned char
unsigned short
short int
unsigned long
I'd pick something based on the ranges you need
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: By default, the int data type is a 32-bit signed two's complement integer, which has a minimum value of -2^31 and a maximum value of 2^31 - 1. In Java SE 8 and later, you can use the int data type to represent an unsigned 32-bit integer, which has a minimum value of 0 and a maximum value of 2^32 - 1. Use the Integer class to use int data type as an unsigned integer. See the section The Number Classes for more information. Static methods like compareUnsigned, divideUnsigned etc have been added to the Integer class to support the arithmetic operations for unsigned integers.
char: The char data type is a single 16-bit Unicode character. It has a minimum value of '\u0000' (or 0) and a maximum value of '\uffff' (or 65,535 inclusive).
long: The long data type is a 64-bit two's complement integer. The signed long has a minimum value of -2^63 and a maximum value of 2^63 - 1. In Java SE 8 and later, you can use the long data type to represent an unsigned 64-bit long, which has a minimum value of 0 and a maximum value of 2^64 - 1. Use this data type when you need a range of values wider than those provided by int. The Long class also contains methods like compareUnsigned, divideUnsigned etc to support arithmetic operations for unsigned long.
short: The short data type is a 16-bit signed two's complement integer. It has a minimum value of -32,768 and a maximum value of 32,767 (inclusive). As with byte, the same guidelines apply: you can use a short to save memory in large arrays, in situations where the memory savings actually matters.
ref: http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
Java doesn't have unsigned types (apart from char, which pretends not to be a number)
byte is one byte, in range -128 to 127
char is two bytes (unsigned), but only generally used for characters, not numbers
short is also two bytes, in range -32768 to 32767
int is four bytes
long is eight bytes
Edit -- char is unsigned (thanks to Hot Licks in comments)
According to docs.oracle.com:-
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.
Byte - 8 bits
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1)
Adding all these numbers we get a total of 255. Then how's the range which is -128 to 127 calculated. Is it hard-coded somewhere or there's some more technicality to this range?
Any suggestions would be appreciated.
Let's calculate range for 1 Byte
1 bit can take 0 or 1
1 Byte = 8 Bits
The first bit is used as a sign ( - or + )
then remaining bits are 7
so we can write 2^7 = 128 different numbers for one sign
we get 0 as a positive sign. then we have 128 numbers for the negative side,127 numbers for the positive side and 0 (zero)
so the range is -128 to 127 including 0
It is a signed type, meaning, it still has a range of 255 (as you have correctly calculated), but it starts at -128. So half the range is below zero, 1 possible number is = (zero) and the remaining 127 are above 0.
The first bit is the sign. (1 - minus, 0 - plus)
Formula for Range calculation is : -2^(n-1) to (2^(n-1)-1)
where n = no. of bits of primitive datatype.
Example:
For int datatype, n is 32, in short datatype, n is 16 etc.
So, int range will be: -2^(32-1) to (2^(32-1)-1)
By using the same formula Range of byte, short, float and double could be calculated.
That's because the first bit is used for the sign, since the data type is signed.
Please refer to http://en.wikipedia.org/wiki/Signed_number_representations
Since there is no unsigned primitive types in Java (like C or C#), is usual to cast it to a bigger type if you need to "overflow" the boundaries.
The last bit i.e. number 8 we are writing it 2^7 is a sign bit that decides negative or positive sign so it is 2^0 +2^1 +2^2 +2^3 +2^4 +2^5+ 2^6
Formula to calculate range in Java
-2(n-1) to +2(n-1)-1
Where n is the number of bit (1 byte= 8 bit)
So, for byte type range would be:
-2(8-1) to +2(8-1)-1
or, -2(7) to +2(7)-1
or, -128 to +127
it results from the standard representation of signed values as binary numbers. For a full description see http://en.wikipedia.org/wiki/Two%27s_complement.
-Rich
Let's calculate range for 1 Byte in C Langauge Datatype char
In case of char dataype lets consider -
1 Byte = 8 Bits
2^8 = 256
so we can write 2^8 = 256
In case of unsigned 0 to positive integers only.
4.1 In case of signed
we have 128 numbers for the negative side, 0 (zero) and 127 numbers for the positive side
so the range is -128 0 +127
Here 0 is taking one bit
Bit consists of 0’s and 1’s.byte normally consists of 8 bits.so the values can be calculated using the general formula which is given below,
no of values data type can have=2^n(2 power n), where n represents no of bits.
so the value of byte data type=2^8(i.e 1 byte=8 bits),here n=8
byte value=256
And it should be shared equal on both sides of zero ( half values at negative and half value at positive ). Hence the range for byte is from -128 to 127.
Range of data types
so now we came to know that how we are calculating the Range of the integer data types.This logic is applicable for all the integer data types.
The full details of all the data types are given below,
S.NO Data Type Bits Ranges values
1 boolean 1 – true or false(1 or 0)
2 byte 8 -128 to 127. 256(2^8)
3 short 16 -32,768 to 32,767 65,536(2^16)
4 int 32 -2^31 to (2^31)-1 2^32
5 long 64 Refer NOTE 2^64
6 float 32 Refer NOTE ———
7 double 64 Refer NOTE ———-
8 char 16 0 to 65,535 65,536(2^16)