max value of integer
JavaCIntegerMaxBitJava Problem Overview
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?
Java Solutions
Solution 1 - Java
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
Solution 2 - Java
> 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.
Solution 3 - Java
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.
Solution 4 - Java
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.
Solution 5 - Java
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.
Solution 6 - Java
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
Solution 7 - Java
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).
Solution 8 - Java
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.
Solution 9 - Java
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.
Solution 10 - Java
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).
Solution 11 - Java
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.