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

Question

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.

Answer 1

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

Answer 2

As others have answered, the "standards" all leave most of the details as "implementation defined" and only state that type "char" is at leat "char_bis" wide, and that "char <= short <= int <= long <= long long" (float and double are pretty much consistent with the IEEE floating point standards, and long double is typically same as double--but may be larger on more current implementations).

Part of the reasons for not having very specific and exact values is because languages like C/C++ were designed to be portable to a large number of hardware platforms--Including computer systems in which the "char" word-size may be 4-bits or 7-bits, or even some value other than the "8-/16-/32-/64-bit" computers the average home computer user is exposed to. (Word-size here meaning how many bits wide the system normally operates on--Again, it's not always 8-bits as home computer users may expect.)

If you really need a object (in the sense of a series of bits representing an integral value) of a specific number of bits, most compilers have some method of specifying that; But it's generally not portable, even between compilers made by the ame company but for different platforms. Some standards and practices (especially limits.h and the like) are common enough that most compilers will have support for determining at the best-fit type for a specific range of values, but not the number of bits used. (That is, if you know you need to hold values between 0 and 127, you can determine that your compiler supports an "int8" type of 8-bits which will be large enought to hold the full range desired, but not something like an "int7" type which would be an exact match for 7-bits.)

Note: Many Un*x source packages used "./configure" script which will probe the compiler/system's capabilities and output a suitable Makefile and config.h. You might examine some of these scripts to see how they work and how they probe the comiler/system capabilities, and follow their lead.

Answer 3

unsigned char bits = sizeof(X) << 3;

where X is a char,int,long etc.. will give you size of X in bits.

Answer 4

For floating point numbers there is a standard (IEEE754): floats are 32 bit and doubles are 64. This is a hardware standard, not a C++ standard, so compilers could theoretically define float and double to some other size, but in practice I've never seen an architecture that used anything different.

Answer 5

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!

Answer 6

On a 64-bit machine:

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