With the aid of pointer-based arithmetic operations and the usage of pointers in comparison operations, address arithmetic is a technique for determining the address of an object. Pointer arithmetic is another name for address arithmetic.
The pointers can be used for mathematical operations like addition, subtraction, etc. The outcome of an arithmetic operation on the pointer, however, will likewise be a pointer if the other operand is of type integer because we know that the pointer includes the address. These operations are addition and subtraction. In C++, a pointer's value can be increased or decreased. It signifies that we can change the pointer's value by adding or removing integer values. A pointer arithmetic can be subtracted (or added) from another in a manner similar to this.
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Answer: True
Explanation:
Yes, the given statement is true that the due to the very low feasibility the IT (Information technology) department are not aware of the hidden backlog that basically contain projects.
The feasibility is the main factor in the IT department as it helps to study whole objective of the project and then also uncover all the strength and weakness of the project so that we can easily go through the details to make the project more efficient and reliable.
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Data erasure software uses standards that are called Typically, a software program device is used for information erasure like BitRaser, which implements the overwriting sample primarily based totally on the usual used, together with US DoD 5220.22, NIST 800-88.
<h3>What is statistics erasure?</h3>
For virtual garage devices, erasing names after its developer, the Gutmann set of rules is a way of disk wiping that overwrites statistics the usage of a complete of 35 passes. This makes it one of the maximum steady statistics erasure methods, however additionally the maximum time-consuming.
The statistics manner that zeros and ones are verifiably overwritten onto all sectors of the device. This renders all statistics completely unrecoverable at the same time as retaining the capability of the device the beyond few years, NIST Special Publication 800-88 has end up the go-to statistics erasure preferred withinside the United States.
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Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
