指针的大小是多少?它具体取决于什么?

What is the size of a pointer? What exactly does it depend on?(指针的大小是多少?它具体取决于什么?)

问题描述

我在网上搜索，虽然可以找到一些讨论，但没有找到全面的描述.因此，如果任何人都可以形成一个涵盖有关指针大小的所有内容的答案，那将有很大帮助.答案至少应涵盖以下主题

I searched online and while I could find a few discussions, I did not find a comprehensive description. So if anyone could form an answer which covers everything about size of a pointer, it would be of great help. The answer should at least cover following topics

1. 指针的大小取决于什么?
2. 架构的什么特性会影响指针的大小?(详细)
3. 编译器如何影响指针的大小?

推荐答案

指针是高级语言提供的抽象；理论上它可以是任何宽度.这完全是编译器的心血来潮.

A pointer is an abstraction provided by a high-level language; in theory it could be any width at all. It's totally at the whim of the compiler.

在实践中，它通常与底层硬件的内存地址的宽度有关，因为这通常是编译器实现的最有效的东西.不过也有例外；例如，C++ 的成员函数指针没有直接映射到硬件地址，因为它需要表示两个实体(函数和类型的一些概念).

In practice, it's typically related to the width of the memory addresses of the underlying hardware, as that's usually the most efficient thing for the compiler to implement. There are exceptions though; for example, C++'s pointer-to-member-function does not have a direct mapping to hardware addresses, as it needs to represent two entities (the function and some notion of the type).

然而，即使撇开这一点，仍然存在复杂性.例如:

However, even leaving that aside, there are still complexities. For example:

• 在大多数现代硬件上，您的程序将使用虚拟内存地址，而不是物理地址(它们的宽度可能不同).除非您正在编写内核空间代码.
• 在某些架构(例如 x86)上，底层硬件展示了一个分段地址空间.这确实很复杂，但大部分都被操作系统和虚拟内存系统抽象掉了.但是，如果您正在为非常老的 x86 编写内核空间代码或代码，则必须处理它.
• 在当前的 x86-64 上，(虚拟)内存地址实际上只有 48 位宽.
• x86-64 支持 32 位和 64 位可执行文件.
• 您可能正在一个虚拟机中运行，该虚拟机可以根据需要执行任何操作(相对于底层物理机).

• On most modern hardware, your program will work with virtual memory addresses, rather than physical addresses (which may not be the same width). Unless you're writing kernel-space code.
• On some architectures (e.g. x86), the underlying hardware exhibits a segmented address space. This is really complicated, but is mostly abstracted away by the OS and the virtual-memory system. If you're writing kernel-space code or code for really old x86s, you'll have to deal with it, though.
• On current x86-64, (virtual) memory addresses are actually only 48-bits wide.
• x86-64 supports both 32-bit and 64-bit executables.
• You may be running inside a virtual machine, which again can do whatever if it wants (relative to the underlying physical machine).

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