百度试题 结果1 题目 Given the IP address 180.25.21.172 and the subnet mask 255.255.192.0. What is the subnet address?( ) A. 180.25.21.0 B. 180.25.0.0 C. 180.25.8.0 D. 180.0.0.0 相关知识点: 试题来源: 解析 b 反馈 收藏
if (line.find("Subnet Mask" != string::npos) { int pos = line.find(": "); if (pos != string::npos) { subnetmask = line.substr(pos+2); break; } } } ... Same way, you can get your own IP address (or use gethostbyname). By using the following function you could check...
Find IP Address from Username. Find IP Address of other clients in my network lan (Remotely)? Find IP address via Username. Find Network ID find the guid for an application in the registry Find windows server license status Finding a Domain Controller Finding Device ID for a given NIC... ...
I just do some head-math to find the 1 multiple of 32 that's less than the number in the Interesting Octet. Then I determine the next-highest multiple because that's the next subnet number. The above tells me that 172.24.240.225/27 is a valid IP host address within the 172.24.240.224...
Given an IP address 172.16.28.252 with a subnet mask of 255.255.240.0, what is the correct networkaddres?()A.172.16.16.0B.176.24.0.0C.172.16.0.0D.172.16.28.0的答案是什么.用刷刷题APP,拍照搜索答疑.刷刷题(shu
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C++ - Set IP address, subnet mask, network gateway C++ - Check given string is numeric or not C++ - Check given date is in valid format or not C++ - Add seconds to the time C++ - Find Fibonacci number C++ - Find next greatest number from the same set of digits C++ - Convert number...
I just happened to leave the config as-is on the router (just added the entry for the access-list to permit the 192.168.1.0 network and changed the IP on FA1 to a 192.168.1.0/24 address.) There also isn't a particular reason that I chose to use a different IP subnet to connect ...
if(maskAddress.AddressFamily!=address.AddressFamily) // We got something like an IPV4-Address for an IPv6-Mask. This is not valid. returnfalse; // Now find out how long the prefix is. intmaskLength=int.Parse(subnetMask.Substring(slashIdx+1)); ...
I just do some head-math to find the 1 multiple of 32 that's less than the number in the Interesting Octet. Then I determine the next-highest multiple because that's the next subnet number. The above tells me that 172.24.240.225/27 is a valid IP host address within the 172.24.240.224...