C进制转换和组合的问题

Question

本帖最后由 jimever 于 2013-10-25 16:58:41 编辑

Answer 1

#include <stdio.h>typedef unsigned char uchar;void swap(uchar *c1, uchar *c2)
{
uchar t = *c1;
*c1 = *c2;
*c2 = t;
}void display(uchar *arr, int len)
{
int i;
for(i = 0; i < len; i++)
printf("0x%02X ", arr[i]);
putchar('\n');
}void displayShort(short *arr, int len)
{
int i;
for(i = 0; i < len; i++)
printf("0x%04X ", arr[i]);
putchar('\n');
}int main()
{
char recv[]={0x35,0x5D,0x03,0xE6}; display(recv, sizeof(recv)); swap(&recv[0], &recv[1]);
swap(&recv[2], &recv[3]); display(recv, sizeof(recv));
displayShort((short*)recv, sizeof(recv)/sizeof(short)); return 0;
}
/*
运行结果:
D:\profile\Desktop\test\c>a
0x35 0x5D 0x03 0xE6
0x5D 0x35 0xE6 0x03
0x355D 0x03E6
*/

Answer 2

#define MAKESHORT(a,b) ((unsigned short)(((LONG)a<<8))|((LONG)b& 0xff))int _tmain(int argc, _TCHAR* argv[])
{
char recv[]={0x35,0x5D,0x03,0xE6};
unsigned short pShort = 0;
char a,b;
for (unsigned int i = 0 ; i < sizeof(recv)/sizeof(short); i++)
{
a = recv[i*2];b = recv[i*2+1];
pShort =  MAKESHORT(a,b);
printf("十进制为 %d  ",pShort);
printf("十六进制 %x\r\n",pShort);
}
getchar();
return 0;
}

Answer 3

写一个联合体不就行了嘛，还不用去考虑大端法和小端法。typedef union _tagUnion
{
short nCombin;
struct 
{
char nFst;
char nSec;
}SEP;}OUT, *POUT;
int main()
{
OUT out;
out.nFst = 0x35;
out.nSec = 0x5D;
printf("%d\n", out.nCombin); out.nFst = 0x03;
out.nSec = 0xE6;
printf("%d\n", out.nCombin); return 0;
}

Answer 4

有个地方写错了：typedef union _tagUnion
{
unsigned short nCombin;
struct 
{
char nFst;
char nSec;
}SEP;}OUT, *POUT;
int main()
{
OUT out;
out.SEP.nFst = 0x35;
out.SEP.nSec = 0x5D;
printf("%d\n", out.nCombin); out.SEP.nFst = 0x03;
out.SEP.nSec = 0xE6;
printf("%d\n", out.nCombin); return 0;
}

Answer 5

char recv[]={0x35,0x5D,0x03,0xE6};short x1=((short)recv[0]<<8)|(recv[1]&0xFF);
short x2=((short)recv[2]<<8)|(recv[3]&0xFF);
printf("%d    %d\n",x1,x2);

Answer 6

方法二:
char   recv[4]={0x35,0x5D,0x03,0xE6};
char   val[4];
int    i;
short  *x=(short *)val;for(i=0;i<4;i+=2)
{
val[i]=recv[i+1];
val[i+1]=recv[i];
}printf("%d    %d\n",*x,*(x+1));