#include <fstream>
#include<iostream>
#include<cstring>
using namespace std;
class Address {
public:
char addr[6];
Address() {}
Address(string address) {
size_t pos = address.find(":");
int id = stoi(address.substr(0, pos));
short port = (short)stoi(address.substr(pos + 1, address.size()-pos-1));
memcpy(addr, &id, sizeof(int));
memcpy(&addr[4], &port, sizeof(short));
}
};
enum MsgTypes{
JOINREQ,
JOINREPLY,
DUMMYLASTMSGTYPE,
HEARTBEAT
};
/**
* STRUCT NAME: MessageHdr
*
* DESCRIPTION: Header and content of a message
*/
typedef struct MessageHdr {
enum MsgTypes msgType;
}MessageHdr;
typedef struct en_msg {
// Number of bytes after the class
int size;
// Source node
Address from;
// Destination node
Address to;
}en_msg;
//class Testing{
void send(Address *myaddr, Address *toaddr, char *data, int size);
int main()
{
MessageHdr *msg=new MessageHdr();
size_t msgsize = sizeof(MessageHdr) + sizeof(Address) + sizeof(long) + 1;
msg=(MessageHdr *)malloc(msgsize*sizeof(char));
int id=233;
short port =22;
long heartbeat=1;
msg=(MessageHdr *)malloc(msgsize*sizeof(char));
string s=to_string(id)+":"+to_string(port);
string s1=to_string(id+1)+":"+to_string(port+1);
cout<<s<<'\n';
cout<<s1<<'\n';
Address *addr= new Address(s);
for (int i = 0; i < 6; i++)
cout << addr->addr[i];
Address *toaddr= new Address(s1);
msg->msgType = JOINREQ;
//cout<<(char *)msg->msgType;
memcpy((char *)(msg+1), addr, sizeof(addr));
memcpy((char *)(msg+1) + 1 + sizeof(addr), &heartbeat, sizeof(long));
send(addr, toaddr, (char *)msg, msgsize);
return 0;
}
void send(Address *myaddr, Address *toaddr, char *data, int size) {
cout<<"inside send"<<'\n';
en_msg *em;
//static char temp[2048];
em = (en_msg *)malloc(sizeof(en_msg) + size);
em->size = size;
memcpy(&(em->from), &(myaddr), sizeof(em->from));
memcpy(&(em->to), &(toaddr), sizeof(em->from));
memcpy(em + 1, data, size);
cout<<(char *)(em+1);
}
这是我的程序,在我之间我试图检查地址存储在我的char数组中。但是在打印阵列时,它会产生一些奇怪的输出。打印s和s1的值后出现两个奇怪的符号。 我试图将id:port存储在地址类的char数组中,但看起来没有成功。请帮忙
我所指的打印代码是主要功能。主函数大约十行。
比如说,我的身份证是233,端口是22,地址是233:22我想要回溯233:22并打印出来。我怎么在这里做到这一点?
提前致谢:)
答案 0 :(得分:1)
问题出在这一行:
cout << addr->addr[i];
由于addr->addr
是char
的数组,因此每个元素都将打印为它所代表的字符。如果您更愿意打印每个的整数值,只需先将其转换为int
。
cout << static_cast<int>(addr->addr[i]); // or old-fashioned: (int)addr->addr[i];
答案 1 :(得分:0)
给出以下代码:
for (int i = 0; i <= 6; i++)
cout << addr->addr[i];
给定Address
的构造函数:
size_t pos = address.find(":");
int id = stoi(address.substr(0, pos));
short port = (short)stoi(address.substr(pos + 1, address.size()-pos-1));
memcpy(addr, &id, sizeof(int));
memcpy(&addr[4], &port, sizeof(short));
很明显,您正在打印符合数字的字节
addr->addr
是一个char数组,它包含两个整数变量,一个有两个字节(int),另一个有2个字节(short)。
所以,如果数字是436,那么你正在打印:
0xB4 0x01 0x00 0x00
<crazy char> SOH NULL NULL
您必须了解要打印的内容或要打印的内容才能正确打印。
注意:这里假设最流行的设置,这意味着:
<强>更新强>
如何获取地址和端口:
int address;
unsigned short port;
memset(&address, addr->addr, 4);
memset(&port, addr->addr+4, 2);