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标题: gethostbyname()的问题 [打印本页]

作者: sprinklexu    时间: 2004-12-09 13:47
标题: gethostbyname()的问题
struct hostent *gethostbyname(const char *name)

传入的参数的格式是什么?比如“www.sohu.com”还是“sohu.com”

传出的参数有没有什么需要注意的问题??
作者: albcamus    时间: 2004-12-09 14:15
标题: gethostbyname()的问题
这个要看DNS怎么解释吧?能解释出来的自然什么格式都行(前提是它们的确是同一个地址),不能的话就返回“Unknow host”。
作者: superdoctor    时间: 2004-12-09 14:47
标题: gethostbyname()的问题
看来楼主对基本的域名知识也没有过关啊,呵

参数肯定是一个域名串啊
"www.sohu.com"的话就返回"www.sohu.com"的A记录
"sohu.com"就返回“sohu.com"的A记录啊

如下:
#nslookup
>;set type=a
>;sohu.com     //在这里输入你需要查询的域名
Server:         202.106.0.20
Address:        202.106.0.20#53

Non-authoritative answer:
Name:   sohu.com
Address: 61.135.150.215    //gethostbyname返回的就是这个
>;www.sohu.com
Server:         202.106.0.20
Address:        202.106.0.20#53

Non-authoritative answer:
www.sohu.com    canonical name = pagegrp1.sohu.com.
Name:   pagegrp1.sohu.com
Address: 61.135.150.75   //如果有多个地址,那么gethostbyname随机返回一个(适用于A记录)
Name:   pagegrp1.sohu.com
Address: 61.135.150.99
Name:   pagegrp1.sohu.com
Address: 61.135.150.101
Name:   pagegrp1.sohu.com
Address: 61.135.150.113
Name:   pagegrp1.sohu.com
Address: 61.135.150.145
Name:   pagegrp1.sohu.com
Address: 61.135.131.73
Name:   pagegrp1.sohu.com
Address: 61.135.131.91
Name:   pagegrp1.sohu.com
Address: 61.135.131.180
Name:   pagegrp1.sohu.com
Address: 61.135.131.182
Name:   pagegrp1.sohu.com
Address: 61.135.131.183
Name:   pagegrp1.sohu.com
Address: 61.135.132.65
Name:   pagegrp1.sohu.com
Address: 61.135.132.80
Name:   pagegrp1.sohu.com
Address: 61.135.132.172
Name:   pagegrp1.sohu.com
Address: 61.135.132.173
Name:   pagegrp1.sohu.com
Address: 61.135.132.176
Name:   pagegrp1.sohu.com
Address: 61.135.133.109
Name:   pagegrp1.sohu.com
Address: 61.135.145.47
Name:   pagegrp1.sohu.com
Address: 61.135.150.65
Name:   pagegrp1.sohu.com
Address: 61.135.150.67
>; quit
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