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.

如果您正在为64位机器编译，那么它可能是8。

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。

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.

据我所知，这是基于内存地址的大小。所以在一个32位地址方案的系统上，sizeof将返回4，因为那是4个字节。

指针大小为4字节的原因是因为您正在为32位体系结构编译。正如FryGuy所指出的，在64位架构上你将看到8。