一、大小端：

1 大端： 數據低位存放在內存高地址， 數據高位存放在內存低地址

2 小端： 數據低位存放在內存底地址， 數據高位存放在內存高地址

3 網絡字節序以大端傳輸、主機字節序根據CPU架構不同既有大端與有小端

以聯合體來做說明：

union endpoint {	unsigned short n;	char ch[2];};int main(){	endpoint endpoint_t;	endpoint_t.n = 0x1122;	if (endpoint_t.ch[0] == 0x22 && endpoint_t.ch[1] == 0x11)		std::cout << "little ending" << std::endl;	else		std::cout << "big ending" << std::endl;
	return 0;}
/*
二、位域已知一個16位的整數，請按4位為一個數， 寫個求和函數。如：16位整數的二進制為1101 0101 1110 0011， 計算結果為 1101 + 0101 + 1110 + 0011 = 35
*/
union endpoint {	unsigned short n;	char ch[2];};typedef struct set_ {#ifndef __LITTLE_ENDING__	unsigned short a : 4;	unsigned short b : 4;	unsigned short c : 4;	unsigned short d : 4;#else	unsigned short c : 4;	unsigned short d : 4;	unsigned short a : 4;	unsigned short b : 4;#endif}set_t;unsigned short binset_plus(const unsigned short n){	set_t s = { 0 };	memcpy(&s, &n, sizeof(short));	return s.a + s.b + s.c + s.d;}void main(){	endpoint endpoint_t;	endpoint_t.n = 0x1122;	if (endpoint_t.ch[0] == 0x22 && endpoint_t.ch[1] == 0x11)		std::cout << "little ending" << std::endl;	else		std::cout << "big ending" << std::endl;	unsigned short n = 0xD5E3;  // 1101 0101 1110 0011	unsigned short na = binset_plus(n);}/*
三、字節對齊
* /
void main()
{	struct node {	};	unsigned short n = sizeof(node);  // 1	struct node1 {		int a;		char b;		short c;	};	n = sizeof(node1);  //因為結構體node1中最長的數據類型是int，占4個字節，因此以4字節對齊，則該結構體在內存中存放方式為						//|--------int--------|						//|char|----|--short--|						// 4 + 4 = 8	struct node2 {		char b;		int a;		short c;	};	n = sizeof(node2);  //因為結構體node2中最長的數據類型是int，占4個字節，因此以4字節對齊，則該結構體在內存中存放方式為						//|-------char--------|						//|--------int--------|						//|-------short-------|						// 4 + 4 + 4= 12	struct node4 {		bool a;		node1 node1_;		short b;	};	n = sizeof(node4); //4 + 8 + 4 因為node1中最寬字段為int a, 占4字節且小于默認的8字節對齊， 所以取4byte為對齊字節	struct node5 {		bool a;		node1 nodde_;		char d;		double b;		char c;	};	n = sizeof(node5);  //8 + 8 + 8 + 8 + 8 = 40#PRagma pack(4)	struct node6 {		bool a;		node1 nodde_;		char d;		double b;		char c;	};	n = sizeof(node6);  //4 + 8 + 4 + 8 + 4 = 28#pragma pack()}