整数的htonl（）和ntohl（）的输出相同 [英] Same output for htonl() and ntohl() on an integer

问题描述

我在低字节序[LE]机器[Linux，Intel处理器]上运行了以下程序。我无法在下面的代码片段中解释这3个输出。由于计算机是LE，因此 a 的值存储为 0x78563412 。打印时，它将显示其实际值。由于它是LE机器，所以我希望 ntohl（）是无操作并显示 0x78563412 在做。但是，我希望包含 htonl（）的第二个打印语句的 0x12345678 。

I ran the following program on little-endian [LE] machine [Linux, Intel processor]. I am unable to explain the 3 outputs in below code snippet. Since machine is LE, the value of a is stored as 0x78563412. When printing, it is displaying its actual value. Since its an LE machine, I expect ntohl() to be a no-op and display 0x78563412, which it is doing. However, I expect 0x12345678 for 2nd print statement containing htonl(). Can someone please help me understand why they are same?

int main() 
{
    int a = 0x12345678; 

    printf("Original - 0x%x\n", (a)); 
    printf("Network - 0x%x\n", htonl(a)); 
    printf("Host - 0x%x\n", ntohl(a)); 

    return 0;
}

输出：

Original - 0x12345678
Network - 0x78563412
Host - 0x78563412

推荐答案

由于它是LE机器，我希望 ntohl（）是一个没有操作

那是错误。网络字节顺序为 big-endian ，主机字节顺序为little-endian。因此， ntohl 和 htonl 均返回其输入的字节交换版本。

That's the mistake. Network byte order is big-endian, host byte order is little-endian. Therefore, both ntohl and htonl return a byte-swapped version of their input.

请记住， htonl 的意义在于您可以在主机上取一个整数，然后输入：

Remember, the point of htonl is that you can take an integer on the host, then write:

int i = htonl(a);

，结果是 i ，当使用网络字节顺序解释时，其值与 a 相同。因此，如果将 i 的对象表示形式写入套接字，并且另一端的阅读器期望网络字节顺序为4字节整数，它将读取的值。 a 。

and the result is that the memory of i, when interpreted using network byte order, has the same value that a does. Hence, if you write the object representation of i to a socket and the reader at the other end expects a 4-byte integer in network byte order, it will read the value of a.

并显示 0x78563412

这是您要写的吗？如果 ntohl 是无操作（或者说是一个身份函数），那么您的第三行必然会打印出与第一行相同的内容，因为您会得到 ntohl（a）== a 。这就是在big-endian实现中发生的情况，您的程序将在其中打印：

Is this what you intended to write? If ntohl were a no-op (or rather, an identity function), then your third line necessarily would print the same thing as your first line, because you would have ntohl(a) == a. This is what happens on big-endian implementations, where your program prints:

Original - 0x12345678
Network - 0x12345678
Host - 0x12345678

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