整数如何存储在内存中？ [英] how are integers stored in memory?

问题描述

我在阅读关于Big / Little Endian的文章时感到困惑。

I'm confused when I was reading an article about Big/Little Endian.

代码如下：

#include <iostream>
using namespace std;

int i = 12345678;

int main()
{
    char *p = (char*)&i;  //line-1

    if(*p == 78)  //line-2
        cout << "little endian" << endl;
    if(*p == 12)
        cout << "big endian" << endl;

}

问题：

1. 在第一行，我可以使用 static_cast< char *>（& i） ？

在第二行，根据代码，如果是小端字符串，则 78 存储在最低字节中，否则 12 存储在最低字节中。但我认为， i = 12345678; 将以二进制的形式存储在内存中。

In line-2, according to the code, if it's little-endian, then 78 is stored in the lowest byte, else 12 is stored in the lowest byte. But what I think is that, i = 12345678; will be stored in memory in binary.

-endian，那么 i 的最后一个字节将存储在最低字节，但是我不明白是如何保证最后一个字节 i 是 78 ？

If it's little-endian, then the last byte of i's binary will be stored in the lowest byte, but what I don't understand is how can it guarantee that the last byte of i is 78?

code> i = 123; ，则 i 的二进制是 01111011 ，可以保证在小端字节中， 23 存储在最低字节中吗？

Just like, if i = 123;, then i's binary is 01111011, can it guarantee that in little-endian, 23 is stored in the lowest byte?

推荐答案

1. 我更喜欢

1. I'd prefer a reinterpret_cast.

Little- endian和big-endian指的是方式字节，即8位数量，存储在存储器中，而不是两位十进制数量。如果 i 的值为 0x12345678 ，则可以检查 0x78 和 0x12 来确定字节顺序，因为两个十六进制数字对应一个字节（在我编程的所有硬件上）。

Little-endian and big-endian refer to the way bytes, i.e. 8-bit quantities, are stored in memory, not two-decimal quantities. If i had the value 0x12345678, then you could check for 0x78 and 0x12 to determine endianness, since two hex digits correspond to a single byte (on all the hardware that I've programmed for).

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