0
1 MB (megabyte) = 1048576 bytes (2^20)
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
How many cylinder is equal to 1bytes?
It depends on how many tracks and sectors, as well as the sector size. In a typical (very old) 360kb floppy, there were 40 cylinders, 2 tracks per cylinder, 9 sectors per track, and 512 bytes per sector so, in that case, 1 byte would be about 1.085 x 10-4 cylinders.
What is 100000000 bytes in GB?
It is either 0.1 GB, or 0.093 GB, depending on whom you ask. Storage devices are advertised as decimal (or "metric") gigabytes, which is 109 or 1,000,000,000 bytes. Microsoft Windows, and indeed most programs written by classic programmers, use binary gigabytes, which is 230 or 1,073,741,824 bytes. The latter definition gives a better view of how much storage is used, because storage devices store sections of 512 (29) bytes at once, not 500 bytes. This eases address computations, etc. as computers naturally use binary numbers. Manufacturers use decimal values so that neophyte computer users can understand exactly how much capacity they're buying.
How do you write a C plus plus code for first fit worst fit best fit algorithms?
The following program shows one method of implementing first-fit, best-fit and worst-fit algorithms. There's a lot of code but it can be broken down into 4 major classes: process, processes, partition and partitions. In the real-world, the partition class would allocate memory to itself and assign one or more processes to that memory (keeping track of which process is where). Likewise, the process class would represent an actual process. However, to keep things simple, we're just modelling these objects to show how the algorithms work in terms of allocating a process to a partition based upon the size of the process and the available bytes in a partition.The process class simply stores an ID and the size of the process (in bytes). The processes class is really nothing more than a wrapper for an embedded vector of processes. Similarly, the partition class holds an ID, the size of the partition and an embedded processes class to keep track of which processes it holds. The partitions class is also nothing more than a wrapper for an embedded vector of partitions.In the main function we can see how these classes are used. First, 5 partitions are allocated sizes of 4096, 2048, 1024, 512 and 256 bytes. In the real-world partitions would be equal size, however we can imagine that they were all 4096 bytes to begin with and that these sizes represent what is currently available. We then create 10 processes from 2048 down to 4 bytes in size. We then randomise the order of both collections to give a more realistic example (the IDs tell us the order in which they were actually created).The key to allocating the processes to the partitions is the processes::allocate function, where we pass a reference to the processes to be allocated, along with the method (first, best or worst).With first-fit, we simply take each process in turn and allocate it to the first partition that is large enough. Since the processes and partitions were randomised to begin with, they could be placed in any order. However, note that the largest process (process 1 with 2048 bytes) can only be placed in partition 1 (4096 bytes) or partition 2 (2048 bytes), but can only be placed in partition 2 if there no other processes occupying that partition. Partition 1 is capable of holding all 10 processes at once, so there should never be the situation where a process cannot be allocated (in the real-world it can happen, in which case you would need to try and add a new partition).With best-fit, we're looking for the partition that leaves the least free bytes for each process. To achieve this we sort the processes in descending order of size. We also sort the partitions in ascending order of size before each process is allocated to a partition. By sorting the collections in this manner, we can re-use the first-fit algorithm, which will always finds the smallest partition that will accept the process.With worst-fit we do the same thing as we did with best-fit, except we reverse the order. Processes are sorted in ascending order of size and partitions in descending order of size. Again, this allows us to use the first-fit algorithm to find the largest partition that will accept the process.In other words, we use the same algorithm for all three, the only difference being the order of the processes and partitions.In the real-world, of course, we'd allocate just one process at a time rather than a collection of processes all at once. Thus the only difference with best-fit and worst-fit is that we do not need to sort the processes at all, but we must still sort the partitions based upon the current available bytes (but not for first-fit). We achieve this by totalling the size of the current processes within that partition and subtract that sum from the partition size. However, in the real world processes may be deallocated in any sequence, which will leave non-contiguous gaps in the partition. We get around that problem by allocating sub-partitions to those gaps, thus every partition is really a tree of sub-partitions that starts out with one node (the root). In order to locate the best or worst fit we must 'flatten' the trees by creating a vector for all the nodes and sub-nodes and then sort that vector by the size of each node. This is well beyond the scope of the answer of course, however the program below contains enough information to be able to apply first, best and worst-fit to any vector, regardless of its source.Program code#include#include#include#include#include// forward declarationsclass process_class;class processes_class;class partition_class;class partitions_class;std::ostream& operator
How many character in 1 KB?
1 Kb is 1024 bytes and 1 character takes 1 byte of the main memory. So, it is 1024 chars in 1 Kb. The preceding is only true for languages which have 8-bit characters. Most modern computer languages support the concept of Unicode, which allows for character encodings in various languages. The most widespread Unicode encoding format is UTF-8, which uses between 1 and 4 bytes to represent a specific character symbol. For instance, the Java programming language assumes all characters are in Unicode UTF-16 format, which is a 16-bit character encoding. So, in Java, only 512 characters will fit in 1 kB.
What is meant by GB data?
GB is acronym for gigabyte and is used to indicate storage capacity on a computer. A gigabyte is 2^30 bytes or 1,073,741,824 bytes. Note that disk manufacturers and memory chip manufacturers use different notations. Memory chips are precisely specified such that a 1 GB memory chip is guaranteed to provide storage for exactly 1,073,741,824 bytes. However, hard drives only give approximate storage capacities such that 1 GB only guarantees storage capacity for at least 10^9 bytes, or 1 billion bytes, which is the equivalent of a 0.93 GB memory chip. Technically, hard-drive manufacturers use the correct notation since the prefixes kilo-, mega- and giga- literally equate to some power-of-ten (10^3, 10^6 and 10^9 respectively). However, it is not possible to create a memory chip with exactly 10^3 bytes, it has to be an exact power-of-two (2^10, 2^20 and 2^30, respectively). The reason memory has to be an exact power-of-two stems from the way in which memory is addressed on a binary computer. In order to address 1000 bytes we'd need at least 10 bits in the range 0000000000 through 1111100111 (0 to 999 decimal) which means there are 24 invalid addresses in the range 1111101000 through 1111111111 (1000 to 1023 decimal). This over-complicates circuit design; it's much easier to make all bit patterns valid by adding on those missing 24 addresses thus giving us the full kilo-binary-byte capacity of 1024 bytes. Hard-drive manufacturers are less constrained in the way storage is addressed, because the addressing is subject to the operating system and the way in which the hard-drive is formatted. Typically, a hard-drive is divided up into addressable clusters which are themselves some power-of-two bytes in length, such as clusters of 512, 1024, 2048 or 4096 bytes. In an effort to avoid confusion, the terms kilo-binary-byte (KiB), mega-binary-byte (MiB) and giga-binary-byte (GiB) are used to indicate precise memory capacities while KB, MB and GB are used to indicate the more generalised hard disk capacities. However, these terms is not widely adopted. For instance, Windows operating systems use the term KB to mean 1024 bytes whether reporting memory capacity or hard drive capacity, hence a 300 GB hard-drive only has capacity for 279.4 GB of storage on Windows. But 279.4 GB is 300 billion bytes so, strictly-speaking, it is quite correct to call it a 300 gigabyte drive.
How many bytes can be stored in one sector?
512
How many bytes is 512 MB?
512 megabytes equates to 536,870,912 bytes.
How many bytes are in 512 kilobytes?
524288 Bytes equals 512 Kilobytes
What can hold up to 512 bytes of data?
A Sector.
How many bytes of data can 32 MB of RAM store?
A stick of 512 megabytes of RAM can hold about 512 megabytes of data. It cannot, however, store it for long because it is volatile and is not designed to store data.
Sectors typically contain how many bytes?
512 bytes
How many bits are in 512 bytes?
4096 bits
How many bytes per sector are there in a Gigabyte?
One Sector on storage media constitutes of 512 Bytes.
A computers has 512 mb memory how many characters can be stored in its memory at a time?
512 x 1024 bytes
If 512 kilo bytes down stream is the recommended speed needed for video calls. What is 512 kilo bytes to mega bytes?
1 mega byte is equal to 1024 kilo bytesso 512 kilo byte turns out to be 0.5 mega bytes
In zone bit recording every sector on the drive has how many bytes?
512
In zone bit recording every sector on the drive has bytes?
512 bytes.
