为什么在C（和C ++）标准中uintptr_t和intptr_t可选类型？ [英] Why are uintptr_t and intptr_t optional types in the C (and C++) standard?

问题描述

对于C99（及更高版本的标准），标头< stdint.h> 中要求某些类型可用。对于精确宽度，例如 int8_t ， int16_t 等，它们是可选的，并且是标准格式

With C99 (and later standards) standard requires certain types to be available in the header <stdint.h>. For exact-width, e.g., int8_t, int16_t, etc..., they are optional and motivated in the standard why that is.

但是对于 uintptr_t 和 intptr_t 类型，也是可选的，但我看不出它们是可选而不是必需的原因。

But for the uintptr_t and intptr_t type, they are also optional but I don't see a reason for them being optional instead of required.

推荐答案

在某些平台上，指针类型具有比任何整数类型都大得多的尺寸。我相信这样的平台示例将是IBM AS / 400，其虚拟指令集将所有指针定义为128位。这种平台的最新示例是Elbrus。它使用128位指针，它们是硬件描述符，而不是普通地址。

On some platforms pointer types have much larger size than any integral type. I believe an example of such as platform would be IBM AS/400 with virtual instruction set defining all pointers as 128-bit. A more recent example of such platform would be Elbrus. It uses 128-bit pointers which are HW descriptors rather than normal addresses.

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