4.6 — Fixed-width integers and size_t

In the previous lessons on integers, we covered that C++ only guarantees that integer variables will have a minimum size -- but they could be larger, depending on the target system.

Why isn’t the size of the integer variables fixed?

The short answer is that this goes back to the early days of C, when computers were slow and performance was of the utmost concern. C opted to intentionally leave the size of an integer open so that the compiler implementers could pick a size for int that performs best on the target computer architecture.

Doesn’t this suck?

By modern standards, yes. As a programmer, it’s a little ridiculous to have to deal with types that have uncertain ranges.

Consider the int type. The minimum size for int is 2 bytes, but it’s often 4 bytes on modern architectures.

If you assume an int is 4 bytes because that’s most likely, then your program will probably misbehave on architectures where int is actually 2 bytes (since you will probably be storing values that require 4 bytes in a 2 byte variable, which will cause overflow or undefined behavior).

If you assume an int is only 2 bytes to ensure maximum compatibility, then the range of values you can safely store in an int is significantly limited. And on systems where int is actually 4 bytes, you’re not making use of half of the memory allocated per int.

Fixed-width integers

To address the above issues, C99 defined a set of fixed-width integers (in the stdint.h header) that are guaranteed to be the same size on any architecture.

These are defined as follows:

C++ officially adopted these fixed-width integers as part of C++11. They can be accessed by including the <cstdint> header, where they are defined inside the std namespace. Here’s an example:

#include <cstdint> // for fixed-width integers
#include <iostream>

int main()
{
    std::int16_t i{5};
    std::cout << i << '\n';
    return 0;
}

The fixed-width integers have two downsides that are typically raised.

First, the fixed-width integers are not guaranteed to be defined on all architectures. They only exist on systems where there are fundamental types matching their widths and following a certain binary representation. Your program will fail to compile on any such architecture that does not support a fixed-width integer that your program is using. However, given that most modern architectures have standardized around 8/16/32/64-bit variables, this is unlikely to be a problem unless your program needs to be portable to some exotic mainframe or embedded architectures.

Second, if you use a fixed-width integer, it may be slower than a wider type on some architectures. For example, if you need an integer that is guaranteed to be 32-bits, you might decide to use std::int32_t, but your CPU might actually be faster at processing 64-bit integers. However, just because your CPU can process a given type faster doesn’t mean your program will be faster overall -- modern programs are often constrained by memory usage rather than CPU, and the larger memory footprint may slow your program more than the faster CPU processing accelerates it. It’s hard to know without actually measuring.

Fast and least integers

To help address the above downsides, C++ also defines two alternative sets of integers that are guaranteed to be defined.

The fast types (std::int_fast#_t and std::uint_fast#_t) provide the fastest signed/unsigned integer type with a width of at least # bits (where # = 8, 16, 32, or 64). For example, std::int_fast32_t will give you the fastest signed integer type that’s at least 32 bits. By fastest, we mean the integral type that can be processed most quickly by the CPU.

The least types (std::int_least#_t and std::uint_least#_t) provide the smallest signed/unsigned integer type with a width of at least # bits (where # = 8, 16, 32, or 64). For example, std::uint_least32_t will give you the smallest unsigned integer type that’s at least 32 bits.

Here’s an example from the author’s Visual Studio (32-bit console application):

#include <cstdint> // for fast and least types
#include <iostream>

int main()
{
	std::cout << "least 8:  " << sizeof(std::int_least8_t) * 8 << " bits\n";
	std::cout << "least 16: " << sizeof(std::int_least16_t) * 8 << " bits\n";
	std::cout << "least 32: " << sizeof(std::int_least32_t) * 8 << " bits\n";
	std::cout << '\n';
	std::cout << "fast 8:  " << sizeof(std::int_fast8_t) * 8 << " bits\n";
	std::cout << "fast 16: " << sizeof(std::int_fast16_t) * 8 << " bits\n";
	std::cout << "fast 32: " << sizeof(std::int_fast32_t) * 8 << " bits\n";

	return 0;
}

This produced the result:

least 8:  8 bits
least 16: 16 bits
least 32: 32 bits

fast 8:  8 bits
fast 16: 32 bits
fast 32: 32 bits

You can see that std::int_least16_t is 16 bits, whereas std::int_fast16_t is actually 32 bits. This is because on the author’s machine, 32-bit integers are faster to process than 16-bit integers.

However, these fast and least integers have their own downsides: First, not many programmers actually use them, and a lack of familiarity can lead to errors. Second, the fast types can lead to memory wastage, as their actual size may be larger than indicated by their name. More seriously, because the size of the fast/least integers can vary, it’s possible that your program may exhibit different behaviors on architectures where they resolve to different sizes. For example:

#include <cstdint>
#include <iostream>

int main()
{
    std::uint_fast16_t sometype { 0 };
    sometype = sometype - 1; // intentionally overflow to invoke wraparound behavior

    std::cout << sometype << '\n';

    return 0;
}

This code will produce different results depending on whether std::uint_fast16_t is 16, 32, or 64 bits.

It’s hard to know where your program might not function as expected until you’ve rigorously tested your program on such architectures.

std::int8_t and std::uint8_t likely behave like chars instead of integers

Due to an oversight in the C++ specification, most compilers define and treat std::int8_t and std::uint8_t (and the corresponding fast and least fixed-width types) identically to types signed char and unsigned char respectively. This means these 8-bit types may (or may not) behave differently than the rest of the fixed-width types, which can lead to errors. This behavior is system-dependent, so a program that behaves correctly on one architecture may not compile or behave correctly on another architecture.

We show an example of this in lesson 4.12 -- Introduction to type conversion and static_cast.

When storing integral values where a specific size is important, it’s generally best to avoid std::int8_t and std::uint8_t (and the related fast and least types) and use std::int16_t or std::uint16_t instead.

Best practices for integral types

Given the various pros and cons of the fundamental integral types, the fixed-width integral types, the fast/least integral types, and signed/unsigned challenges, there is little consensus on integral best practices.

Our stance is that it’s better to be correct than fast, and better to fail at compile time than runtime. Therefore, if you need an integral type with a fixed size, we recommend avoiding the fast/least types in favor of the fixed-width types. If you later discover the need to support a platform for which a specific fixed-width type won’t compile, then you can decide how to migrate your program (and thoroughly retest) at that point.

What is std::size_t?

Consider the following code:

#include <iostream>

int main()
{
    std::cout << sizeof(int) << '\n';

    return 0;
}

On the author’s machine, this prints:

4

Pretty simple, right? We can infer that operator sizeof returns an integer value -- but what integral type is that return value? An int? A short? The answer is that sizeof returns a value of type std::size_t. std::size_t is an alias for an implementation-defined unsigned integral type. In other words, the compiler decides if std::size_t is an unsigned int, an unsigned long, an unsigned long long, etc…

std::size_t is defined in a number of different headers. If you need to use std::size_t, <cstddef> is the best header to include, as it contains the least number of other defined identifiers.

For example:

#include <cstddef>  // for std::size_t
#include <iostream>

int main()
{
    int x { 5 };
    std::size_t s { sizeof(x) }; // sizeof returns a value of type std::size_t, so that should be the type of s
    std::cout << s << '\n';

    return 0;
}

Much like an integer can vary in size depending on the system, std::size_t also varies in size. std::size_t is guaranteed to be unsigned and at least 16 bits, but on most systems will be equivalent to the address-width of the application. That is, for 32-bit applications, std::size_t will typically be a 32-bit unsigned integer, and for a 64-bit application, std::size_t will typically be a 64-bit unsigned integer.

std::size_t imposes an upper limit on the size of an object created on the system Optional

Amusingly, we can use the sizeof operator (which returns a value of type std::size_t) to ask for the size of std::size_t itself:

#include <cstddef> // for std::size_t
#include <iostream>

int main()
{
	std::cout << sizeof(std::size_t) << '\n';

	return 0;
}

Compiled as a 32-bit (4 byte) console app on the author’s system, this prints:

4

sizeof must be able to return the byte-size of an object as a value of type std::size_t. Therefore, the byte-size of an object can be no larger than the largest value std::size_t can hold.

The C++20 standard ([basic.compound] 1.8.2) says: “Constructing a type such that the number of bytes in its object representation exceeds the maximum value representable in the type std::size_t (17.2) is ill-formed.”

If it were possible to create a larger object, sizeof would not be able to return its byte-size, as it would be outside the range that a std::size_t could hold. Creating an object with a size (in bytes) larger than the largest value an object of type std::size_t can hold is therefore invalid (and will cause a compile error).

For example, let’s assume that std::size_t has a size of 4 bytes on our system. An unsigned 4-byte integral type has range 0 to 4,294,967,295. Therefore, a 4-byte std::size_t object can hold any value from 0 to 4,294,967,295. Any object with a byte-size of 0 to 4,294,967,295 could have it’s size returned in a value of type std::size_t, so this is fine. However, if the byte-size of an object were larger than 4,294,967,295 bytes, then sizeof would not be able to return the size of that object accurately, as the value would be outside the range of a std::size_t. Therefore, no object larger than 4,294,967,295 bytes could be created on this system.

When 8-bit and 16-bit applications were the norm, this imposed a significant constraint on the size of objects. In the 32-bit and 64-bit era, this is rarely an issue, and therefore not something you generally need to worry about.