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Reptile [31]
3 years ago
10

Given the following code, what is output by the method call, mystery(6 * 8)? public static void mystery (int x[]) { System.out.p

rintln("A"); } public static void mystery (int x) { System.out.println("B"); } public static void mystery (String x) { System.out.println("C"); }
Computers and Technology
1 answer:
____ [38]3 years ago
7 0

Answer:

B

Explanation:

This question demonstrates the concept of method overloading

Method overloading means having different versions of the same method. In this case the method mystery() has three versions. The compiler is able to determine which method to call by their different parameters

The first implementation of mystery requires an array of integers as parameter

The second implementation requires an int variable

The third implementation requires a String

The Method call mystery(6 * 8) will output B because 6*8 evaluates to an integer which is the expected argument.

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void Print_3_Element_Subsets(int n,const element_type S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

print("No subset found");

}

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

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}

}

}

}

C++ Implementation;

#include<cstdlib>

#include<time.h>

using namespace std;

int S1[5]= {1,2,3,4,5};

int n1 = 5;

int S2[2]={1,2};

int n2 = 2;

void Print_3_Element_Subsets(int n,const int S[])

{

if(n<3) // condition to check if list have less than 3 elements

{

printf("No subset found\n");

}

printf("3 Subsets are\n");

for(int i =0; i < n ; i++)

{

for(int j =i+1; j < n ; j++)

{

for(int k =j+1; k < n ; k++)

{

cout<<S[i]<<" "<<S[j]<<" "<<S[k]<<endl;

}

}

}

}

int main()

{

cout<<"Case 1"<<endl;

Print_3_Element_Subsets(n1,S1);

cout<<endl<<"Case 2"<<endl;

Print_3_Element_Subsets(n2,S2);

}

OUTPUT

Case 1

3 Subsets are

1 2 3

1 2 4

1 2 5

1 3 4

1 3 5

1 4 5

2 3 4

2 3 5

2 4 5

3 4 5

Case 2

No subset found

3 Subsets are

--------------------------------

Process exited after 0.0137 seconds with return value 0

Press any key to continue . . .

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