The size of the pointer basically depends on the architecture of the system in which it is implemented. For example the size of a pointer in 32 bit is 4 bytes (32 bit ) and 8 bytes(64 bit ) in a 64 bit machines. The bit types in a machine are nothing but memory address, that it can have. 32 bit machines can have 2^32 address space and 64 bit machines can have upto 2^64 address spaces. So a pointer (variable which points to a memory location) should be able to point to any of the memory address (2^32 for 32 bit and 2^64 for 64 bit) that a machines holds.

由于这个原因，我们看到指针的大小在32位机器中是4字节，在64位机器中是8字节。

你得到的保证是sizeof(char) == 1。没有其他保证，包括没有保证sizeof(int *) == sizeof(double *)。

在实践中，指针的大小在16位系统上为2(如果您能找到一个)，在32位系统上为4，在64位系统上为8，但是依赖于给定的大小并不能获得任何东西。

在windows 32位机器上的Turbo C编译器中，指针和int的大小为2字节。

所以指针的大小是编译器特定的。但一般大多数编译器实现为支持32位4字节指针变量和64位8字节指针变量)。

所以指针的大小在所有机器上都是不一样的。

指针只是一个地址的容器。在32位计算机上，您的地址范围是32位，因此指针总是4字节。在64位机器上，如果你的地址范围是64位，一个指针将是8字节。

一般来说，当你在不同的平台上编译时，sizeof(几乎任何东西)都会改变。在32位平台上，指针的大小总是相同的。在其他平台上(64位是最明显的例子)，这种情况可能会发生变化。

从技术上讲，C标准只保证sizeof(char) == 1，其余的取决于实现。但在现代x86架构(例如Intel/AMD芯片)上，这是相当可预测的。

You've probably heard processors described as being 16-bit, 32-bit, 64-bit, etc. This usually means that the processor uses N-bits for integers. Since pointers store memory addresses, and memory addresses are integers, this effectively tells you how many bits are going to be used for pointers. sizeof is usually measured in bytes, so code compiled for 32-bit processors will report the size of pointers to be 4 (32 bits / 8 bits per byte), and code for 64-bit processors will report the size of pointers to be 8 (64 bits / 8 bits per byte). This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte, to address more memory you need integers larger than 32-bits.