Address 1) House number and street name Address 2) City or town Address 3) State and/or county Address 4) Zip or postal code Address 5) Country [optional]
An IPv4 address has four groups of digits from 0 to 255 (256 total in one group which is equal 2^8), so it could vary from 0.0.0.0 to 255.255.255.255 Total length of IPv4 address space is 2^32 = 4,294,967,296 total addresses. An IPv6 address has eight groups of four hexadecimal digits from 0 to ffff (65,536 total in each group which is equal 2^16), so it could vary from 0:0:0:0:0:0:0:0 to ffff:ffff:ffff:ffff:ffff:ffff:ffff:ffff Total length of IPv6 address space is 2^128 = 340,282,366,920,938,463,463,374,607,431,768,211,456 total addresses, which means it's more than 4,8 * 10^28 of addresses for each of 7 billion people on the planet.
To find the number of divisors of a number, factor it into primes in power format; then the number of factors the number has is the product of the powers of the primes each incremented by 1. To find the proper divisors subtract one from thin number. The prime factorization of 7920 is 2^4 x 3^2 x 5 x 11, so it has (4 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 5 × 3 × 2 × 2 = 60 factors. As this includes the number itself and the proper factors of a number exclude the number, 7920 has 60 - 1 = 59 proper factors (divisors).
2^32 is amount of blocks that address bus could locate. and each blocks is 64bit because data bus has 64 lines. then maximum number of bits stored in memory is (2^32)*64 bit. By: Mohammad Saghafi Email: mohammads1364@yahoo.com
- x - 2 = 1 - 2xAdd 2x to each side:x - 2 = 1Add 2 to each side:x = 3
The top of stack to copied to the specified register and the stack pointer is incremented by 2. A special form of POP, RET, has the program continuing with the popped address in the program counter, i.e. a return from subroutine or function call.
1. find out the system base address from memory map of system. for 8259a it is FF00H. 2. find the internal address for the device. in 8259A there are 2 internal pin A0 and A1. 3.add the internal address to the bus address to get the system address for each internal address.
There are 2 main IP protocols. In IP version 4, each IP address has 4 octets. In IP version 6, each IP address has 16 octets.If somebody says "IP address" without further qualifications, he probably means IP version 4, since that is the current standard. IP version 6 is the planned future standard.There are 2 main IP protocols. In IP version 4, each IP address has 4 octets. In IP version 6, each IP address has 16 octets.If somebody says "IP address" without further qualifications, he probably means IP version 4, since that is the current standard. IP version 6 is the planned future standard.There are 2 main IP protocols. In IP version 4, each IP address has 4 octets. In IP version 6, each IP address has 16 octets.If somebody says "IP address" without further qualifications, he probably means IP version 4, since that is the current standard. IP version 6 is the planned future standard.There are 2 main IP protocols. In IP version 4, each IP address has 4 octets. In IP version 6, each IP address has 16 octets.If somebody says "IP address" without further qualifications, he probably means IP version 4, since that is the current standard. IP version 6 is the planned future standard.
Its purely based on the ip address of each process. You can contact your administrator or you can get the ISP ip address from the link
Specific or static? Each computer has an address, most of the time a TCP/IP address in a pattern of four sets of three numbers. There's also MAC address, which is a pattern of 8 sets of 2 digits, which are assigned to every network adapter.
A memory address a, is said to be n-byte aligned when a is a multiple of n bytes (where n is a power of 2). In this context a byte is the smallest unit of memory access, i.e. each memory address specifies a different byte.
ANSWER There are 2128 combinations of addresses. This is about 3.4 x 1038 locations. Assuming each address holds a 32-bit word, that's 1.2 x 1039 bytes. That's a LOT of memory.
Recall that paging is implemented by breaking up an address into a page and offset number. It is most efficient to break the address into X page bits and Y offset bits, rather than perform arithmetic on the address to calculate the page number and offset. Because each bit position represents a power of 2, splitting an address between bits results in a page size that is a power of 2.
If there are 2^n cells then you need at least n bits to address each cell individually. For example, if there 2^32 cells then you need at least 32 bits to address each of the cells. 2^32 = 4,294,967,296 cells The cells are numbered the range 0 to 4,294,967,295 (inclusive). In hexadecimal notation this range becomes 0h to FFFFFFFFh. The upper bound has exactly 8 hex digits. Each hex digit represents the 4-bit binary value 1111. 8 x 4 bits = 32 bits. QED.
Try thisFor Each nic As System.Net.NetworkInformation.NetworkInterface In System.Net.NetworkInformation.NetworkInterface.GetAllNetworkInterfaces()If nic.OperationalStatus = Net.NetworkInformation.OperationalStatus.Up ThenMessageBox.Show(String.Format("The MAC address of {0} is{1}{2}", nic.Description, Environment.NewLine, nic.GetPhysicalAddress()))End IfNextAnil
32 bit address line can access 4GB of memory. As 2^10 -> 1KB; 2^20 -> 2MB; 2^30 -> 1GB and so on.... 32 bit gives (2^30) * (2^2) = 1GB * 4 = 4GB;
This refers to switching at layer 2 of the OSI reference model, for example, Ethernet. A switch looks at the MAC address of each Ethernet frame ("packet", you might say, but at this level the correct name is "frame"), and if it knows that this MAC address is connected at a certain port, the switch will send the information out ONLY through that port.This refers to switching at layer 2 of the OSI reference model, for example, Ethernet. A switch looks at the MAC address of each Ethernet frame ("packet", you might say, but at this level the correct name is "frame"), and if it knows that this MAC address is connected at a certain port, the switch will send the information out ONLY through that port.This refers to switching at layer 2 of the OSI reference model, for example, Ethernet. A switch looks at the MAC address of each Ethernet frame ("packet", you might say, but at this level the correct name is "frame"), and if it knows that this MAC address is connected at a certain port, the switch will send the information out ONLY through that port.This refers to switching at layer 2 of the OSI reference model, for example, Ethernet. A switch looks at the MAC address of each Ethernet frame ("packet", you might say, but at this level the correct name is "frame"), and if it knows that this MAC address is connected at a certain port, the switch will send the information out ONLY through that port.