c++ - how - long long int range




What does the C++ standard state the size of int, long type to be? (16)

1) Table N1 in article "The forgotten problems of 64-bit programs development"

2) "Data model"

I'm looking for detailed information regarding the size of basic C++ types. I know that it depends on the architecture (16 bits, 32 bits, 64 bits) and the compiler.

But are there any standards for C++?

I'm using Visual Studio 2008 on a 32-bit architecture. Here is what I get:

char  : 1 byte
short : 2 bytes
int   : 4 bytes
long  : 4 bytes
float : 4 bytes
double: 8 bytes

I tried to find, without much success, reliable information stating the sizes of char, short, int, long, double, float (and other types I didn't think of) under different architectures and compilers.


As mentioned the size should reflect the current architecture. You could take a peak around in limits.h if you want to see how your current compiler is handling things.



For 32-bit systems, the 'de facto' standard is ILP32 — that is, int, long and pointer are all 32-bit quantities.

For 64-bit systems, the primary Unix 'de facto' standard is LP64 — long and pointer are 64-bit (but int is 32-bit). The Windows 64-bit standard is LLP64 — long long and pointer are 64-bit (but long and int are both 32-bit).

At one time, some Unix systems used an ILP64 organization.

None of these de facto standards is legislated by the C standard (ISO/IEC 9899:1999), but all are permitted by it.

And, by definition, sizeof(char) is 1, notwithstanding the test in the Perl configure script.

Note that there were machines (Crays) where CHAR_BIT was much larger than 8. That meant, IIRC, that sizeof(int) was also 1, because both char and int were 32-bit.


From Alex B The C++ standard does not specify the size of integral types in bytes, but it specifies minimum ranges they must be able to hold. You can infer minimum size in bits from the required range. You can infer minimum size in bytes from that and the value of the CHAR_BIT macro that defines the number of bits in a byte (in all but the most obscure platforms it's 8, and it can't be less than 8).

One additional constraint for char is that its size is always 1 byte, or CHAR_BIT bits (hence the name).

Minimum ranges required by the standard (page 22) are:

and Data Type Ranges on MSDN:

signed char: -127 to 127 (note, not -128 to 127; this accommodates 1's-complement platforms) unsigned char: 0 to 255 "plain" char: -127 to 127 or 0 to 255 (depends on default char signedness) signed short: -32767 to 32767 unsigned short: 0 to 65535 signed int: -32767 to 32767 unsigned int: 0 to 65535 signed long: -2147483647 to 2147483647 unsigned long: 0 to 4294967295 signed long long: -9223372036854775807 to 9223372036854775807 unsigned long long: 0 to 18446744073709551615 A C++ (or C) implementation can define the size of a type in bytes sizeof(type) to any value, as long as

the expression sizeof(type) * CHAR_BIT evaluates to the number of bits enough to contain required ranges, and the ordering of type is still valid (e.g. sizeof(int) <= sizeof(long)). The actual implementation-specific ranges can be found in header in C, or in C++ (or even better, templated std::numeric_limits in header).

For example, this is how you will find maximum range for int:

C:

#include <limits.h>
const int min_int = INT_MIN;
const int max_int = INT_MAX;

C++:

#include <limits>
const int min_int = std::numeric_limits<int>::min();
const int max_int = std::numeric_limits<int>::max();

This is correct, however, you were also right in saying that: char : 1 byte short : 2 bytes int : 4 bytes long : 4 bytes float : 4 bytes double : 8 bytes

Because 32 bit architectures are still the default and most used, and they have kept these standard sizes since the pre-32 bit days when memory was less available, and for backwards compatibility and standardization it remained the same. Even 64 bit systems tend to use these and have extentions/modifications. Please reference this for more information:

http://en.cppreference.com/w/cpp/language/types


I notice that all the other answers here have focused almost exclusively on integral types, while the questioner also asked about floating-points.

I don't think the C++ standard requires it, but compilers for the most common platforms these days generally follow the IEEE754 standard for their floating-point numbers. This standard specifies four types of binary floating-point (as well as some BCD formats, which I've never seen support for in C++ compilers):

  • Half precision (binary16) - 11-bit significand, exponent range -14 to 15
  • Single precision (binary32) - 24-bit significand, exponent range -126 to 127
  • Double precision (binary64) - 53-bit significand, exponent range -1022 to 1023
  • Quadruple precision (binary128) - 113-bit significand, exponent range -16382 to 16383

How does this map onto C++ types, then? Generally the float uses single precision; thus, sizeof(float) = 4. Then double uses double precision (I believe that's the source of the name double), and long double may be either double or quadruple precision (it's quadruple on my system, but on 32-bit systems it may be double). I don't know of any compilers that offer half precision floating-points.

In summary, this is the usual:

  • sizeof(float) = 4
  • sizeof(double) = 8
  • sizeof(long double) = 8 or 16

If you need fixed size types, use types like uint32_t (unsigned integer 32 bits) defined in stdint.h. They are specified in C99.


In practice there's no such thing. Often you can expect std::size_t to represent the unsigned native integer size on current architecture. i.e. 16-bit, 32-bit or 64-bit but it isn't always the case as pointed out in the comments to this answer.

As far as all the other built-in types go, it really depends on the compiler. Here's two excerpts taken from the current working draft of the latest C++ standard:

There are five standard signed integer types : signed char, short int, int, long int, and long long int. In this list, each type provides at least as much storage as those preceding it in the list.

For each of the standard signed integer types, there exists a corresponding (but different) standard unsigned integer type: unsigned char, unsigned short int, unsigned int, unsigned long int, and unsigned long long int, each of which occupies the same amount of storage and has the same alignment requirements.

If you want to you can statically (compile-time) assert the sizeof these fundamental types. It will alert people to think about porting your code if the sizeof assumptions change.


On a 64-bit machine:

int: 4
long: 8
long long: 8
void*: 8
size_t: 8

The C++ Standard says it like this:

3.9.1, §2:

There are five signed integer types : "signed char", "short int", "int", "long int", and "long long int". In this list, each type provides at least as much storage as those preceding it in the list. Plain ints have the natural size suggested by the architecture of the execution environment (44); the other signed integer types are provided to meet special needs.

(44) that is, large enough to contain any value in the range of INT_MIN and INT_MAX, as defined in the header <climits>.

The conclusion: It depends on which architecture you're working on. Any other assumption is false.


There are four types of integers based on size:

  • short integer: 2 byte
  • long integer: 4 byte
  • long long integer: 8 byte
  • integer: depends upon the compiler (16 bit, 32 bit, or 64 bit)

There is a standard and it is specified in the various standards documents (ISO, ANSI and whatnot).

Wikipedia has a great page explaining the various types and the max they may store: Integer in Computer Science.

However even with a standard C++ compiler you can find out relatively easily using the following code snippet:

#include <iostream>
#include <limits>


int main() {
    // Change the template parameter to the various different types.
    std::cout << std::numeric_limits<int>::max() << std::endl;
}

Documentation for std::numeric_limits can be found at Roguewave. It includes a plethora of other commands you can call to find out the various limits. This can be used with any arbitrary type that conveys size, for example std::streamsize.

John's answer contains the best description, as those are guaranteed to hold. No matter what platform you are on, there is another good page that goes into more detail as to how many bits each type MUST contain: int types, which are defined in the standard.

I hope this helps!


Updated: C++11 brought the types from TR1 officially into the standard:

  • long long int
  • unsigned long long int

And the "sized" types from <cstdint>

  • int8_t
  • int16_t
  • int32_t
  • int64_t
  • (and the unsigned counterparts).

Plus you get:

  • int_least8_t
  • int_least16_t
  • int_least32_t
  • int_least64_t
  • Plus the unsigned counterparts.

These types represent the smallest integer types with at least the specified number of bits. Likewise there are the "fastest" integer types with at least the specified number of bits:

  • int_fast8_t
  • int_fast16_t
  • int_fast32_t
  • int_fast64_t
  • Plus the unsigned versions.

What "fast" means, if anything, is up to the implementation. It need not be the fastest for all purposes either.


We are allowed to define a synonym for the type so we can create our own "standard".

On a machine in which sizeof(int) == 4, we can define:

typedef int int32;

int32 i;
int32 j;
...

So when we transfer the code to a different machine where actually the size of long int is 4, we can just redefine the single occurrence of int.

typedef long int int32;

int32 i;
int32 j;
...

You can use variables provided by libraries such as OpenGL, Qt, etc.

For example, Qt provides qint8 (guaranteed to be 8-bit on all platforms supported by Qt), qint16, qint32, qint64, quint8, quint16, quint32, quint64, etc.


You can use:

cout << "size of datatype = " << sizeof(datatype) << endl;

datatype = int, long int etc. You will be able to see the size for whichever datatype you type.





c++-faq