Answer:
Websphere Studio Application Developer
Solution:
The process of transaction can guarantee the reliability of business applications. Locking resources is widely used in distributed transaction management (e.g; two phase commit, 2PC) to keep the system consistent. The locking mechanism, however, potentially results in various deadlocks. In service oriented architecture, the deadlock problem becomes even worse because multiple transactions try to lock shared resources in the unexpectable way due to the more randomicity of transaction requests, which has not been solved by existing research results. In this paper, we investigate how to prevent local deadlocks, caused by the resource competition among multiple sub-transactions of a gl obal transaction, and global deadlocks from the competition among different global transactions. We propose a replication based approach to avoid the local deadlocks, and a timestamp based approach to significantly mitigate the global deadlocks. A general algorithm is designed for both local and global deadlock prevention. The experimental results demonstrate the effectiveness and efficiency of our deadlock prevention approach. Further, it is also proved that our approach provides higher system performance than traditional resource allocation schemes.
This is the required answer.
<span>Olfactory memory tends to be the most durable. Surprisingly, smells tend to be the most strongly related with memories. This has been shown by people being able to associate smells with events for longer periods than any other sensory input. Most other inputs only stay in memory for a period of milliseconds, up to 2-4 seconds at the most.</span>
Answer:
it should be d
Explanation:
please mark this as brainlists i could really use it
Answer:
"The value of the variable will remain the same which is already have when the sub-processor is called".
Explanation:
The above question said that:-
void fun(int a)
{
a=a+1;
}
void main()
{
int a=5;
fun(a);
}
//what will be the value of a in the main function after the fun function is excuted.
- Then the answer is: the value of a will be 5 in the main function.
- It is because when the fun function is called, then a variable that is defined in the fun function is a local variable for fun function. That scope after the fun function is null.
- The a variable inside the fun function is a different variable and the main function a variable is also a different variable.
- So when the user prints the value of a variable inside the fun function, it will give the result as 6.
- But when he prints the value of a variable inside the main function, then it will give the value as 5.
