Create
方法应产生以这种方式组成的9个字符的字符串:som
后跟id
字符加4个字符(代表value
的十六进制字符串)加上2个字符(代表crc的十六进制字符串)和eom字符。这是代码:
private:
char som = '$';
char eom = '&';
bool crc = true;
unsigned char checksum(string data) {
unsigned char sum = 0x00;
for (char &c : data)
sum += c;
return sum;
}
string epilogue(string data) {
ostringstream _message;
if (crc) {
byte _sum = checksum(data);
_message << std::hex << (_sum & 0xFF);
}
_message << eom;
return _message.str();
}
string Create(char id, short value) {
ostringstream _message, _data;
_data << std::hex << (value & 0xFFFF);
_message << som << id << _data.str() << epilogue(_data.str());
return _message.str();
}
问题在此行:
_data << std::hex << (value & 0xFFFF);
可以正确输出十六进制字符串,但长度不能为correct。类似的问题应该在该行中:
_message << std::hex << (_sum & 0xFF);
如何解决此代码以产生所需的模式?在Java中,我使用的是这样的内容:
String.format("%02X", value & 0xFFFF)
编辑(可运行代码):
#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
using namespace std;
typedef unsigned char byte;
class PacketBuilder {
private:
char som = '$';
char eom = '&';
bool crc = true;
public:
PacketBuilder(const char som, const char eom, bool crc) {
this->som = som;
this->eom = eom;
this->crc = crc;
}
byte checksum(string data) {
byte sum = 0x00;
for (char &c : data)
sum += c;
return sum;
}
string epilogue(string data) {
ostringstream _message;
if (crc) {
byte _sum = checksum(data);
_message << std::hex << (_sum & 0xFF);
}
_message << eom;
return _message.str();
}
string Create(char id, short value) {
ostringstream _message, _data;
_data << std::hex << (value & 0xFFFF);
_message << som << id << _data.str() << epilogue(_data.str());
return _message.str();
}
};
int main() {
PacketBuilder* _pb = new PacketBuilder('$','&',true);
string _packet = _pb->Create('C',0);
cout << _packet;
return 0;
}
预期结果是:
$C0000c0&
在C ++中使用std::setw
和std::setfill
。
记住要保存和恢复iomanip标志。