如何使用htonl将Little Endian转换为Big Endian

问题描述

我具有以下元素的结构。

I have struct with the following elements. Plus the structure is complety padded.

typedef struct {
    uint16_t a;
    uint16_t b;
    uint8_t c;
    uint8_t d;
    uint8_t e[6];
} ad;

此结构有点小尾数。我的意思是，当我在大字节序机器上打印此结构时，如果c = 1，d = 2，e [0] = 3，e [1]，则得到以下结果

This structure is a little endian. I mean when I print this structure on my big endian machine I get the following

] = 4。
我得到

if c=1 , d=2, e[0] =3, e[1]=4.I get

c = 4，d = 3，e [0] = 2和e [1] = 1。

c=4, d=3, e[0] = 2 and e[1]=1.

a和b被交换。
，e [1]与c交换，e [0]与d交换。

a and b are swapped.further, e[1] is swapped with c and e[0] is swapped with d.

我正在使用 htonl 函数如下所示。

I am using htonl function like the following. but, it is not working, can anyone suggest me a good answer.

推荐答案

Endian-ness仅适用于单个字段，而不适用结构字段的顺序。在您列出的字段中，仅定义为uint16_t的多字节整数字段将由字节顺序控制，而uint8_t是单字节，因此不存在排序问题。无论字节顺序如何，单字节值数组都将保持相同的长度。

Endian-ness only applies to individual fields, not the order of struct fields. Of the fields you list, only the multi-byte integer fields defined as uint16_t will are governed by endian-ness the uint8_t are single byte, so there is no ordering issue to consider. The array of single byte values will also maintain the same length regardless of the endian-ness.

要转换uint16_t，您将需要使用htons（）函数来代替htonl（）。 htonl（）函数期望的长度通常至少为4个字节。

To convert the uint16_t, you will want to use the htons() function instead of htonl(). The htonl() function expects a long which will typically be at least 4 bytes.

uint16_t netShort = htons(uint16_t hostShort);

或者以您的示例struct：

Or for your example struct:

struct.a = htons(struct.a);
struct.b = htons(struct.b);

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