我在Win32中发送实际二进制位作为十六进制。代码是
int binText[32]={1,0,1,0,1,1,1,1,0,0,0,1,1,0,1,0,1,1,1,1,0,1,0,1,1,1,1,1,0,0,0,1}; //hex: AF 1A F5 F1
char tempInt[255]={0};
for (int i=0; i<32; i++)
{
sprintf(&tempInt[strlen(tempInt)], "%d", binText[i]);
}
char HexBuffer[255];
unsigned long int Number = 0;
int BinLength = strlen(tempInt);
for(int i=0; i<32; i++)
{
Number += (long int)((tempInt[32 - i - 1] - 48) * pow((double)2, i));
}
//When i send using following code it sends F1 F5 1A AF
//serialObj.send((char *)&Number);
//So it is changed into big endian using following code.
unsigned char c1, c2, c3, c4;
c1 = Number & 255;
c2 = (Number >> 8) & 255; //>>right shift operator
c3 = (Number >> 16) & 255;
c4 = (Number >> 24) & 255;
unsigned long int Number1= ((int)c1 << 24) + ((int)c2 << 16) + ((int)c3 << 8) + c4;
//and now following code sends AF 1A F5 F1
serialObj.send((char *)&Number1);
现在我必须在AF 1A F5 F1前面附加三个十六进制字节24 24 3F。我们如何将这些十六进制字节附加到Number1。
“serialObj.send()”调用的“send”函数如下:
void serial::send(char data[])
{
DWORD dwBytesWrite;
WriteFile(serialHandle, data, 7, &dwBytesWrite, NULL);
}
答案 0 :(得分:2)
不是将指向32位变量的指针强制转换为char *
,为什么不将这些东西组合成字符呢?
char buf[7] = { 0x24, 0x24, 0x3F }; // first three bytes are prefix
如果你真的需要将一个0和1的32-int数组转换为剩下的四个字节,那么下面的循环将更加简洁:
char buf[7] = { 0x24, 0x24, 0x3F }; // first three bytes are prefix
unsigned char byte = 0;
for (int i = 0; i < 32; i++)
{
byte = (byte << 1) | binText[i];
if ( (i & 7) == 7 )
{
buf[ 3 + (i >> 3) ] = byte; // The "3 +" part adds new bytes past the header
byte = 0;
}
}
serialObj.send(buf);
但是,如果字符串AF 1A F5 F1
实际上是固定的,那么您只需要:
char buf[7] = { 0x24, 0x24, 0x3F, 0xAF, 0x1A, 0xF5, 0xF1 };
serialObj.send(buf);