Program size is not related to system memory usage. Final Fantasy VII for the PC, for instance, needed 1.5 GB of space, yet could run in just 32 MB of RAM. By comparison, NetBeans (an integrated development environment) requires 815 MB of space and 512MB of RAM.
Not sure what you mean; if you want to measure the "input size" in bytes, that would probably be 8 bytes, since integers typically use 4 bytes.
The minimum size of an Ethernet frame is 64 bytes. Even if the VLAN tag is 4 bytes, the Ethernet frame with VLAN tagging remains 64 bytes.
Bytes of memory are chucks of code that help put a program or file together. The largest we have so far is petabyte then going down in size are: terabyte, gigabyte, megabyte, kilobyte, then just byte, then even smaller is a bit.
1 megabyte = 1 048 576 bytes
8 bytes
Totally depends on the hardware and software and their MTU (Maximum Transmission Unit). Usually MTU is set to 1500 bytes.
Short Name = Full name = Size in bytes = Size in bits KB = Kilobyte = 1000 Bytes = 8000 Bits Kb = Kilobit = 125 Bytes = 1000 Bits in previous answer there is mistake kb=12.5 bytes
2048 bytes
The storage size of an int in C is loosely defined, and may be either 2 bytes or, more commonly, 4 bytes. Whether or not it is defined as const won't affect the size.
An Ethernet frame typically has a maximum payload size of 1500 bytes, which includes the APR (Address Resolution Protocol) packet. The APR packet itself is relatively small, usually around 28 bytes, allowing the Ethernet frame to accommodate additional headers and trailers, such as the Ethernet header (14 bytes) and the Frame Check Sequence (4 bytes). Therefore, the total size of an Ethernet frame carrying an APR packet would be at least 64 bytes (minimum frame size) up to 1518 bytes (maximum frame size).
The block size specifies the minimum amount of space that will be used when writing disk blocks. The example, if you write 300 bytes of information to the disk and your block size for the file system is 512 bytes, then 512 bytes (a block) will be written to the disk. The first 300 bytes will be used and the rest to fill out the block of 512 bytes will not be used. This allows for efficient space utilization (assuming the block size is reasonable) and for retrieval of information from the storage device.
Bytes is a unit of size not time so there is no comparison. rubiconn.com