64位计算机上int和sizeof int指针的大小 [英] Size of int and sizeof int pointer on a 64 bit machine
问题描述
我只是想知道我怎么知道我的笔记本电脑是64位还是32位机器。 （这是64）。
I was just wondering how can I know if my laptop is 64 or 32 bit machine. (it is a 64).
所以，我考虑打印以下内容：
So, I thought about printing the following:
int main()
{
printf("%d",sizeof(int));
}
结果是4，这看起来很奇怪（因为它是64位机器）
and the result was 4, which seemed weird (since it is a 64 bit machine)
但是，当我打印出来时：
But, when I printed this:
int main()
{
printf("%d",sizeof(int*));
}
结果是8，这更有意义。
the result was 8, which made more sense.
问题是：
因为我使用的是64位机器，所以不应该使用原始类型如int应该使用8 bytes
Since I'm using a 64 bit machine, shouldn't a primitive type such as int should use 8 bytes
（64位），那个sizeof int应该是8？为什么不是这样？
(64 bit) and by that sizeof int should be 8? Why isn't it so?
为什么int * size是8？
And why is the int* size is 8?
有点困惑在这里，
所以提前感谢。
推荐答案
不， sizeof（int）是实现定义的，通常是4个字节。
No, the sizeof(int) is implementation defined, and is usually 4 bytes.
另一方面，为了解决超过4GB的内存（32位系统可以做到的），你需要你的指针是8字节宽。 int * 只是将地址保存到内存中的某个地方，而你只能用32位来处理超过4GB的内存。
On the other hand, in order to address more than 4GB of memory (that 32bit systems can do), you need your pointers to be 8 bytes wide. int* just holds the address to "somewhere in memory", and you can't address more than 4GB of memory with just 32 bits.
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