C语言 如何从 getaddrinfo() 打印 IP 地址
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How to print IP address from getaddrinfo()
提问by user1345414
I'm trying to use socket in C on CentOS 6.4.
我正在尝试在 CentOS 6.4 上使用 C 中的套接字。
Following LIST1 is my code.
下面的 LIST1 是我的代码。
My code gets hostname from command line and sends datagram to server with UDP successfully.
我的代码从命令行获取主机名并使用 UDP 成功将数据报发送到服务器。
What I want to know is how to print IP address that getaddrinfo()resolved wiht 192.168.10.1format.
我想知道的是如何打印用格式getaddrinfo()解析的IP地址192.168.10.1。
When I try to print IP address segmentation error happens.
当我尝试打印 IP 地址分段错误时发生。
Does anyone know how to fix this code?
有谁知道如何修复此代码？
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <netdb.h>
#include <errno.h>
int
main(int argc,char *argv[]){
int sock;
struct addrinfo hints,*res;
int n;
int err;
if(argc != 2){
fprintf(stderr,"Usage : %s dst \n",argv[0]);
return 1;
}
memset(&hints,0,sizeof(hints));
hints.ai_family = AF_UNSPEC;
hints.ai_socktype = SOCK_DGRAM;
err = getaddrinfo(argv[1],"12345",&hints,&res);
if(err != 0){
perror("getaddrinfo");
printf("getaddrinfo %s\n",strerror(errno));
printf("getaddrinfo : %s \n",gai_strerror(err));
return 1;
}
sock = socket(res->ai_family,res->ai_socktype,0);
if(sock < 0){
perror("socket");
return 1;
const char *ipverstr;
switch (res->ai_family){
case AF_INET:
ipverstr = "IPv4";
break;
case AF_INET6:
ipverstr = "IPv6";
break;
default:
ipverstr = "unknown";
break;
}
printf("ipverstr = %s\n ",ipverstr);
}
n = sendto(sock,"HELLO",5,0,res->ai_addr,res->ai_addrlen);
if(n<1){
perror("sendto");
return 1;
}
struct sockaddr_in *addr;
addr = (struct sockaddr_in *)res->ai_addr;
printf("inet_ntoa(in_addr)sin = %s\n",inet_ntoa((struct in_addr)addr->sin_addr));
printf("############ finish !! #######\n");
close(sock);
freeaddrinfo(res);
return 0;
}
回答by alk
The code misses to include the prototype for inet_ntoa().
代码没有包含inet_ntoa().
The compiler should have told you this.
编译器应该告诉你这一点。
Do add:
添加：
#include <arpa/inet.h>
However the code still compiles as due to the missing protoype for inet_ntoa()it is assumed to return int, whereas it returns a char*which is a pointer, which on a 64bit system is 8 bytes, which is not the same size as intwhich typically has a size of 4. Due to this mismatch things go terribly wrong and end up in a segmentation violation.
然而，由于缺少原型，代码仍然可以编译，因为inet_ntoa()它被假定为 return int，而它返回char*一个指针，它在 64 位系统上是 8 个字节，它的大小与int通常具有4. 由于这种不匹配，事情会变得非常错误并最终导致分段违规。
Also please note: inet_ntoa()is to be used for ipv4 addresses only. Verbatim from man inet_ntoa(italics by me):
另请注意：inet_ntoa()仅用于 ipv4 地址。逐字来自man inet_ntoa（我用斜体）：
The inet_ntoa()function converts the Internet host address in, given in network byte order, to a string in IPv4dotted-decimal notation. The string is returned in a statically allocated buffer, which subsequent calls will overwrite
该 INET_NTOA（）函数将在，互联网主机地址在网络字节顺序给出，在一个字符串的IPv4点分十进制格式。字符串在静态分配的缓冲区中返回，后续调用将覆盖该缓冲区
To be able to convert both (IPv4 and IPv6) struct sockaddr_XYZ's binary addresses to a char[]use inet_ntop().
为了能够将（IPv4 和 IPv6）struct sockaddr_XYZ的二进制地址都转换为char[]使用inet_ntop().
回答by glglgl
The counterpart of getaddrinfo()is getnameinfo(). This turns a struct sockaddrinto a string.
的对应物getaddrinfo()是getnameinfo()。这会将 astruct sockaddr转换为字符串。
