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 <=
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
long int in C.
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
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
long depends on the compiler implementation.
E.g. on my Ubuntu 64 bit I have
32 bits, when on another one 32bit Ubuntu version it is
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’.
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).