Answer:
printStackTrace()
Explanation:
printStackTrace() :- This method is present in Java.lang.Throwable class and it prints this Throwable it also prints other details with throwable like backtrace and class name. printStackTrace() prints a stack trace for this Throwable object on the output stream of standard error.
So we conclude that the answer is printStackTrace().
Linear probing
It does a linear search for an empty slot when a collision is identified
Advantages
Easy to implement
It always finds a location if there is one
Disadvantages
When clusters and keys fill most of the array after forming in adjacent slots of the table, the performance deteriorates
Double probing/hashing
The idea here is to make the offset to the next position probed depending on the key value. This is so it can be different for different keys.
Advantages
Good for double number generation
Smaller hash tables can be used.
Disadvantages
As the table fills up, the performance degrades.
Quadratic probing
It is used to resolve collisions in hash tables. It is an open addressing scheme in computer programming.
Advantage
It is more efficient for a closed hash table.
Disadvantage
Has secondary clustering. Two keys have same probe sequence when they hash to the same location.
Answer:
The expression to this question can be defined as follows:
Expression:
(x >= 0 && y<0) //check condition using AND operator
Explanation:
In the given question it is defined that x and y are an integer variable that holds some value, in which it defines a condition that checks the value of variable x is positive, and the value of y is negative.
- To check positive value a condition is used, that checks x greater than equal to 0, and to check negative value, it uses condition, that is y is less than 0.
- In the above condition, the AND operator is used, which executes when both conditions are true.
Answer:
D
Explanation:
because you need people to access software through hardware and data through information and procedures.
int sum = 0, n;
do {cin>>n; sum+=n;}while (n!=0);
cout<<sum;
