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inna [77]
3 years ago
6

Overloading in methods are popular in programming, and why overloading is important.

Computers and Technology
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

 Yes, overloading is one of the methods which are popular in programming language. Overloading basically refers to the same function but different signature called function overloading or method overloading. It is the ability to define the multiples method by using the single identifier.

The overloading is important because it has the ability to design the multiple method by using similar name. It also provide the high flexibility to the programmers to call the same method in the data. overloading basically provide the high clarity in the code.

Overloading is used to achieved the compile time polymorphism.  

Following are program of function overloading in c++ are:

Class abc // creating class

{

public:

int p;

void fun() // function fun with no parameter/

{

cout<<” hello “;

}

void fun(int a) // function fun with parameter

{

p=a;

cout<<p;

}

};

int main() // main function

{

abc ob; // creating object

ob.fun();// print hello;

ob.fun(6);// print 6

return 0;

}

Explanation:

In this program the function fun() have same name but different signature in the main method we create the object of class abc i.e ob. ob.fun() this statement called the function with no parameter and ob.fun(6) this statement will called the function with integer parameter.

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A Windows user has been successfully saving documents to the file server all morning. However, the latest attempt resulted in th
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c. The share to the file server is disconnected.

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Virtualization:
Sedaia [141]

Answer:

C. can boost server utilization rates to 70% or higher.

Explanation:

Virtualization -

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3 years ago
A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slo
krok68 [10]

Answer:

The answer is below

Explanation:

Given that:

Frame transmission time (X) = 40 ms

Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request

a) Probability that there is success on the first attempt = e^{-G}G^k but k = 0, therefore Probability that there is success on the first attempt = e^{-G}=e^{-2}=0.135

b) probability of exactly k collisions and then a success = P(collisions in k attempts) × P(success in k+1 attempt)

P(collisions in k attempts) = [1-Probability that there is success on the first attempt]^k = [1-e^{-G}]^k=[1-0.135]^k=0.865^k

P(success in k+1 attempt) = e^{-G}=e^{-2}=0.135

Probability of exactly k collisions and then a success = 0.865^k0.135

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3 years ago
F
Alex_Xolod [135]

Answer:

everything

A.

B.

C.

NOOO.

3 0
2 years ago
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