什么是 uint_fast32_t，为什么要使用它而不是常规的 int 和 uint32_t？

Question

So the reason for typedef :ed primitive data types is to abstract the low-level representation and make it easier to comprehend ( uint64_t instead of long long type, which is 8 bytes).

However, there is uint_fast32_t which has the same typedef as uint32_t . Will using the "fast" version make the program faster?

Answer 1

• int may be as small as 16 bits on some platforms. It may not be sufficient for your application.
• uint32_t is not guaranteed to exist. It's an optional typedef that the implementation must provide iff it has an unsigned integer type of exactly 32-bits. Some have a 9-bit bytes for example, so they don't have a uint32_t .
• uint_fast32_t states your intent clearly: it's a type of at least 32 bits which is the best from a performance point-of-view. uint_fast32_t may be in fact 64 bits long. It's up to the implementation.

... there is uint_fast32_t which has the same typedef as uint32_t ...

What you are looking at is not the standard. It's a particular implementation (BlackBerry). So you can't deduce from there that uint_fast32_t is always the same as uint32_t .

See also:

• Exotic architectures the standards committees care about .

• My opinion-based pragmatic view of integer types in C and C++ .

Answer 2

The difference lies in their exact-ness and availability.

The doc here says:

unsigned integer type with width of exactly 8, 16, 32 and 64 bits respectively ( provided only if the implementation directly supports the type ):

 uint8_t uint16_t uint32_t uint64_t

And

fastest unsigned unsigned integer type with width of at least 8, 16, 32 and 64 bits respectively

uint_fast8_t uint_fast16_t uint_fast32_t uint_fast64_t

So the difference is pretty much clear that uint32_t is a type which has exactly 32 bits, and an implementation should provide it only if it has type with exactly 32 bits, and then it can typedef that type as uint32_t . This means, uint32_t may or may not be available .

On the other hand, uint_fast32_t is a type which has at least 32 bits, which also means, if an implementation may typedef uint32_t as uint_fast32_t if it provides uint32_t . If it doesn't provide uint32_t , then uint_fast32_t could be a typedef of any type which has at least 32 bits.

Answer 3

When you #include inttypes.h in your program, you get access to a bunch of different ways for representing integers.

The uint_fast*_t type simply defines the fastest type for representing a given number of bits.

Think about it this way: you define a variable of type short and use it several times in the program, which is totally valid. However, the system you're working on might work more quickly with values of type int . By defining a variable as type uint_fast*t , the computer simply chooses the most efficient representation that it can work with.

If there is no difference between these representations, then the system chooses whichever one it wants, and uses it consistently throughout.

Answer 4

Note that the fast version could be larger than 32 bits. While the fast int will fit nicely in a register and be aligned and the like: but, it will use more memory. If you have large arrays of these your program will be slower due to more memory cache hits and bandwidth.

I don't think modern CPUS will benefit from fast_int32, since generally the sign extending of 32 to 64 bit can happen during the load instruction and the idea that there is a 'native' integer format that is faster is old fashioned.