什么是uint_fast32_t，为什么应使用它代替常规的int和uint32_t？ [英] What is uint_fast32_t and why should it be used instead of the regular int and uint32_t?

问题描述

所以使用 typedef ：ed原始数据类型的原因是抽象了低级表示形式并使其易于理解（ uint64_t 而不是 long long 类型，即8个字节）。

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).

但是，有 uint_fast32_t 具有相同的 typedef 为 uint32_t 。

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

推荐答案

• 在某些平台上，int 可能小到16位。


• uint32_t 不保证存在。
实现必须提供一个可选的 typedef ，前提是该实现必须具有准确的32位无符号整数类型。例如，有些文件有9位字节，因此它们没有 uint32_t 。

• uint_fast32_t 清楚地表明您的意图：至少是 32位类型，从性能的角度来看这是最好的。 uint_fast32_t 实际上可能是64位长。


• 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.




...有 uint_fast32_t code>与 uint32_t 具有相同的typedef ...

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

您正在寻找的不是标准。这是一种特殊的实现方式（BlackBerry）。因此，您不能从那里推断 uint_fast32_t 始终与 uint32_t 相同。

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.

另请参见：




• 标准委员会关注的异国架构。

我基于意见的C和C ++中整数类型的实用视图。

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