The question makes no sense. What do you mean "physical space"? The most straightforward interpretation is that you want to know how big a one-byte storage device would be, and the answer to that is that it depends on what the storage device is (in a complicated way, so don't bother specifying "a hard drive" like that's going to help).
4 bytes are equivalent to 32 bits.
The 8086 can address 1,114,080 bytes. (One Mb + 64Kb - 16) That does not count I/O space, it only counts memory space.
1,000,000,000,000,000,000,000 bytes are equal to 1 Zettabyte. There are also 1 billion terabytes in 1 Zettabyte.
Two gigabytes (GB) is equivalent to 2 x 1,073,741,824 bytes, since one gigabyte is 1,073,741,824 bytes. Therefore, two gigabytes contain 2,147,483,648 bytes.
A Mac address is a 48bit addressing scheme (usually represented in HEX). There are 8 bits in a bytes therefore it is 6 bytes long.
A MAC address is typically 48 bits in length, which is equivalent to 6 bytes. Since each byte consists of 8 bits, a MAC address occupies 6 bytes in total.
1073741824 bytes or 10243 bytes or 230 bytes
1024 bytes
536870912 Bytes
1024 bytes
how many bytes are needed to structure PCB Also explain different purposes of these bytes
In 32 bit address space it will most likely be 4 bytes, since 8 bits is a byte and 32 bits / 8 bits = 4. In 64 bit address space it should be 8 bytes (64 bits / 8 bits = 8). It is architecture dependent so use the sizeof() function.