1个字节，2个字节，4个字节数据之间赋值

1，多个字节的数据赋值给单个字节

#include<stdio.h>typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;int main(void)
{printf("u32 is %d byte(s)\t u16 is %d byte(s)\t u8 is %d byte(s)\n",sizeof(u32),sizeof(u16),sizeof(u8));u32 intNum = 0x11223344;u16 shortNum = 0x5566;u8  charNum = 0x77;charNum = intNum;printf("The charNum is %x.\t", charNum); charNum = shortNum;printf("Now the charNum is %x.\t", charNum);return 0;}

运行结果：

结论：多字节数据作为左值赋给单字节的右值时，由于单字节的“容量”不足以存放多字节，数据截断到可以存放为止，目前的标准是存放最低位到足够位置存放。所以u32的0x11223344赋给u8的charNum时，相较截断了高位的0x112233,把0x44赋给了charNum;同理0x5566的0x55也被截断，只留低位的0x66被赋给charNum。

2，单字节数组的数据赋值给多字节数据

也可以把u32的数据看成是4个u8数据组成的数组，u16看成2个u8

#include<stdio.h>typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;int main(void)
{int i = 0;u32 intNum = 0x11223344;u16 shortNum = 0x5566;u8  charNum[4] = {0x77,0x88,0x99,0xAA};u8  charNum1[2] = { 0xbb,0xcc };/* output the primary data of array charNum and charNum1 */printf("output the primary data of array charNum and charNum1\n");for ( i = 0;i < sizeof(charNum);i++){printf("The charNum[%d] is %x.",i, charNum[i]);}printf("\n");for (i = 0;i < sizeof(charNum1);i++){printf("The charNum1[%d] is %x.", i, charNum1[i]);}printf("\n\n");/* output the shifted data of array charNum and charNum1 */printf("output the shifted data of array charNum and charNum1\n");charNum[0] = ((u32)intNum>>24);     //0x11  = 0x00000011 = 0x11223344>>24   charNum[1] = ((u32)intNum >> 16);   //0x22  = 0x00001122 = 0x11223344>>16  charNum[2] = ((u32)intNum >> 8);    //0x33  = 0x00112233 = 0x11223344>>8   charNum[3] =       intNum;          //0x44  = 0x11223344 = 0x11223344for (i = 0;i < sizeof(charNum);i++){printf("The charNum[%d] is %x.", i, charNum[i]);}printf("\n");charNum1[0] = ((u16)shortNum >> 8); //0x55 = 0x0055 = 0x5566>>8 charNum1[1] = shortNum;      //0x66 = 0x5566for (i = 0;i < sizeof(charNum1);i++){printf("The charNum[%d] is %x.", i, charNum1[i]);}printf("\n");return 0;}

运行结果：

结论：多字节数据赋值给等值单字节数组时，会使用右移将多字节数据移到最低位字节，就像这样下面这句，移位后打印的charNum[0]为0x11;

charNum[0] = ((u32)intNum>>24); //0x11223344>>24  0x00000011

当移位后除最低位字节以外的字节仍然有非0数据时也不用担心，它会因为内存不够的关系被截断，而只剩下最低位字节数据。例如这句,移位后打印的charNum[2]为0x33。

charNum[2] = ((u32)intNum >> 8);  //0x11223344>>8 0x00112233

3，多字节数据赋值给单字节数组的数据

#include<stdio.h>typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;int main(void)
{int i = 0;u32 intNum = 0x11223344;u16 shortNum = 0x5566;u8  charNum[4] = {0x77,0x88,0x99,0xAA};u8  charNum1[2] = { 0xbb,0xcc };/* output the primary data of intNum */printf("output the primary data of charNum and charNum1\n");printf("The intNum is %x, shortNum is %x.", intNum, shortNum);printf("\n\n");/* output the shifted data of intNum */printf("output the shifted data of intNum and shortNum\n");intNum = ((u32)charNum[0] << 24) | ((u32)charNum[1] << 16) | ((u32)charNum[2] << 8) | charNum[3];
//0x77000000 = 0x00000077<<24  0x77880000 = 0x00000088<<16  0x77889900 = 0x00000099<<8 0x778899AA = 0x000000AA<<24  printf("The intNum is %x\n", intNum);shortNum = ((u16)charNum1[0] << 8) + charNum1[1];//0xBB00 = 0x00BB<<8     0xBBCC = 0x00CCprintf("The shortNum is %x\n", shortNum);return 0;}

运行结果：

结论：单字节数组赋值给多字节数据时，采用左移，将单字节数据移到多字节对应位的位置。另外上面的例程也说明了对数组的运算，“+”和“|”运算机制不一样，但结果是一样的。因为目标操作位的位置正好是0，所以结果一样，如果目标操作位有数据时，则会体现2种运算的差异，例如

printf("%d  %d   %d   %d ", (2 + 3), (2 | 3), (2 | 8), (2 + 8));

运行结果：

PS:如果用多个表达式进行移位赋值，像这样

#include<stdio.h>typedef unsigned char u8;
typedef unsigned short u16;
typedef unsigned int u32;int main(void)
{int i = 0;u32 intNum = 0x11223344;u16 shortNum = 0x5566;u8  charNum[4] = {0x77,0x88,0x99,0xAA};u8  charNum1[2] = { 0xbb,0xcc };/* output the primary data of intNum */printf("output the primary data of charNum and charNum1\n");printf("The intNum is %x, shortNum is %x.", intNum, shortNum);printf("\n\n");/* output the shifted data of intNum */printf("output the shifted data of intNum and shortNum\n");/*   expect                            real                       */intNum = ((u32)charNum[0] << 24); //0x77000000 = 0x00000077<<24        0x77000000 = 0x00000077<<24intNum = ((u32)charNum[1] << 16);  //0x77880000 = 0x00000088<<16        0x00880000 = 0x00000088<<16intNum = ((u32)charNum[2] << 8);   //0x77889900 = 0x00000099<<8     0x00009900 = 0x00000099<<8intNum =       charNum[3];            //0x778899AA = 0x000000AA<<24        0x000000AA = 0x000000AA<<24//intNum = ((u32)charNum[0] << 24) | ((u32)charNum[1] << 16) | ((u32)charNum[2] << 8)| charNum[3];printf("The intNum is %x\n", intNum);/*   expect                            real                        */shortNum = ((u16)charNum1[0] << 8);       //0xBB00 = 0x00BB<<8              0xBB00 = 0x00BB<<8shortNum = charNum1[1];                  //0xBBCC = 0x00CC                0x00CC = 0x00CC//shortNum = ((u16)charNum1[0] << 8) + charNum1[1];printf("The shortNum is %x\n", shortNum);return 0;}

运行结果：

结论：得到目标数据只是最后一次赋值的结果，而不是4（2）次赋值的总和。