What's the difference between "int" and "int_fast16_t"?

Question

As I understand it, the C specification says that type int is supposed to be the most efficient type on target platform that contains at least 16 bits.

Isn't that exactly what the C99 definition of int_fast16_t is too?

Maybe they put it in there just for consistency, since the other int_fastXX_t are needed?

Update

To summarize discussion below:

• My question was wrong in many ways. The C standard does not specify bitness for int. It gives a range [-32767,32767] that it must contain.
• I realize at first most people would say, "but that range implies at least 16-bits!" But C doesn't require two's-compliment storage of integers. If they had said "16-bit", there may be some platforms that have 1-bit parity, 1-bit sign, and 14-bit magnitude that would still being "meeting the standard", but not satisfy that range.
• The standard does not say anything about int being the most efficient type. Aside from size requirements above, int can be decided by the compiler developer based on whatever criteria they deem most important. (speed, size, backward compatibility, etc)
• On the other hand, int_fast16_t is like providing a hint to the compiler that it should use a type that is optimum for performance, possibly at the expense of any other tradeoff.
• Likewise, int_least16_t would tell the compiler to use the smallest type that's >= 16-bits, even if it would be slower. Good for preserving space in large arrays and stuff.

Example: MSVC on x86-64 has a 32-bit int, even on 64-bit systems. MS chose to do this because too many people assumed int would always be exactly 32-bits, and so a lot of ABIs would break. However, it's possible that int_fast32_t would be a 64-bit number if 64-bit values were faster on x86-64. (Which I don't think is actually the case, but it just demonstrates the point)

Answer 1

int is a "most efficient type" in speed/size - but that is not specified by per the C spec. It must be 16 or more bits.

int_fast16_t is most efficient type in speed with at least the range of a 16 bit int.

Example: A given platform may have decided that int should be 32-bit for many reasons, not only speed. The same system may find a different type is fastest for 16-bit integers.

Example: In a 64-bit machine, where one would expect to have int as 64-bit, a compiler may use a mode with 32-bit int compilation for compatibility. In this mode, int_fast16_t could be 64-bit as that is natively the fastest width for it avoids alignment issues, etc.

Answer 2

int_fast16_t is guaranteed to be the fastest int with a size of at least 16 bits. int has no guarantee of its size except that:

 sizeof(char) = 1 and sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long).

And that it can hold the range of -32767 to +32767.

(7.20.1.3p2) "The typedef name int_fastN_t designates the fastest signed integer type with a width of at least N. The typedef name uint_fastN_t designates the fastest unsigned integer type with a width of at least N."

Answer 3

As I understand it, the C specification says that type int is supposed to be the most efficient type on target platform that contains at least 16 bits.

Here's what the standard actually says about int: (N1570 draft, section 6.2.5, paragraph 5):

A "plain" int object has the natural size suggested by the architecture of the execution environment (large enough to contain any value in the range INT_MIN to INT_MAX as defined in the header <limits.h>).

The reference to INT_MIN and INT_MAX is perhaps slightly misleading; those values are chosen based on the characteristics of type int, not the other way around.

And the phrase "the natural size" is also slightly misleading. Depending on the target architecture, there may not be just one "natural" size for an integer type.

Elsewhere, the standard says that INT_MIN must be at most -32767, and INT_MAX must be at least +32767, which implies that int is at least 16 bits.

Here's what the standard says about int_fast16_t (7.20.1.3):

Each of the following types designates an integer type that is usually fastest to operate with among all integer types that have at least the specified width.

with a footnote:

The designated type is not guaranteed to be fastest for all purposes; if the implementation has no clear grounds for choosing one type over another, it will simply pick some integer type satisfying the signedness and width requirements.

The requirements for int and int_fast16_t are similar but not identical -- and they're similarly vague.

In practice, the size of int is often chosen based on criteria other than "the natural size" -- or that phrase is interpreted for convenience. Often the size of int for a new architecture is chosen to match the size for an existing architecture, to minimize the difficulty of porting code. And there's a fairly strong motivation to make int no wider than 32 bits, so that the types char, short, and int can cover sizes of 8, 16, and 32 bits. On 64-bit systems, particularly x86-64, the "natural" size is probably 64 bits, but most C compilers make int 32 bits rather than 64 (and some compilers even make long just 32 bits).

The choice of the underlying type for int_fast16_t is, I suspect, less dependent on such considerations, since any code that uses it is explicitly asking for a fast 16-bit signed integer type. A lot of existing code makes assumptions about the characteristics of int that go beyond what the standard guarantees, and compiler developers have to cater to such code if they want their compilers to be used.

Answer 4

The difference is that the fast types are allowed to be wider than their counterparts (without fast) for efficiency/optimization purposes. But the C standard by no means guarantees they are actually faster.

C11, 7.20.1.3 Fastest minimum-width integer types

1 Each of the following types designates an integer type that is usually fastest 262) to operate with among all integer types that have at least the specified width.

2 The typedef name int_fastN_t designates the fastest signed integer type with a width of at least N. The typedef name uint_fastN_t designates the fastest unsigned integer type with a width of at least N.

262) The designated type is not guaranteed to be fastest for all purposes; if the implementation has no clear grounds for choosing one type over another, it will simply pick some integer type satisfying the signedness and width requirements.

Another difference is that fast and least types are required types whereas other exact width types are optional:

3 The following types are required: int_fast8_t int_fast16_t int_fast32_t int_fast64_t uint_fast8_t uint_fast16_t uint_fast32_t uint_fast64_t All other types of this form are optional.

Answer 5

From the C99 rationale 7.8 Format conversion of integer types <inttypes.h> (document that accompanies with Standard), emphasis mine:

C89 specifies that the language should support four signed and unsigned integer data types, char, short, int and long, but places very little requirement on their size other than that int and short be at least 16 bits and long be at least as long as int and not smaller than 32 bits. For 16-bit systems, most implementations assign 8, 16, 16 and 32 bits to char, short, int, and long, respectively. For 32-bit systems, the common practice is to assign 8, 16, 32 and 32 bits to these types. This difference in int size can create some problems for users who migrate from one system to another which assigns different sizes to integer types, because Standard C’s integer promotion rule can produce silent changes unexpectedly. The need for defining an extended integer type increased with the introduction of 64-bit systems.

The purpose of <inttypes.h> is to provide a set of integer types whose definitions are consistent across machines and independent of operating systems and other implementation idiosyncrasies. It defines, via typedef, integer types of various sizes. Implementations are free to typedef them as Standard C integer types or extensions that they support. Consistent use of this header will greatly increase the portability of a user’s program across platforms.

The main difference between int and int_fast16_t is that the latter is likely to be free of these "implementation idiosyncrasies". You may think of it as something like:

I don't care about current OS/implementation "politics" of int size. Just give me whatever the fastest signed integer type with at least 16 bits is.

Answer 6

On some platforms, using 16-bit values may be much slower than using 32-bit values [e.g. an 8-bit or 16-bit store would require performing a 32-bit load, modifying the loaded value, and writing back the result]. Even if one could fit twice as many 16-bit values in a cache as 32-bit values (the normal situation where 16-bit values would be faster than 32-bit values on 32-bit systems), the need to have every write preceded by a read would negate any speed advantage that could produce unless a data structure was read far more often than it was written. On such platforms, a type like int_fast16_t would likely be 32 bits.

That having been said, the Standard does not unfortunately allow what would be the most helpful semantics for a compiler, which would be to allow variables of type int_fast16_t whose address is not taken to arbitrarily behave as 16-bit types or larger types, depending upon what is convenient. Consider, for example, the method:

int32_t blah(int32_t x)
{
  int_fast16_t y = x;
  return y;
}

On many platforms, 16-bit integers stored in memory can often be manipulated just as those stored in registers, but there are no instructions to perform 16-bit operations on registers. If an int_fast16_t variable stored in memory are only capable of holding -32768 to +32767, that same restriction would apply to int_fast16_t variables stored in registers. Since coercing oversized values into signed integer types too small to hold them is implementation-defined behavior, that would compel the above code to add instructions to sign-extend the lower 16 bits of x before returning it; if the Standard allowed for such a type, a flexible "at least 16 bits, but more if convenient" type could eliminate the need for such instructions.

Answer 7

An example of how the two types might be different: suppose there’s an architecture where 8-bit, 16-bit, 32-bit and 64-bit arithmetic are equally fast. (The i386 comes close.) Then, the implementer might use a LLP64 model, or better yet allow the programmer to choose between ILP64, LP64 and LLP64, since there’s a lot of code out there that assumes long is exactly 32 bits, and that sizeof(int) <= sizeof(void*) <= sizeof(long). Any 64-bit implementation must violate at least one of these assumptions.

In that case, int would probably be 32 bits wide, because that will break the least code from other systems, but uint_fast16_t could still be 16 bits wide, saving space.