Answer:
d) either e1 or e5
Explanation:
Here, the instruction i1 goes ahead in trying to open the given file through an input stream buffer reader. If the given file name is wrong, it will indicate that an e1 file is not found or if any other IO errors due to invalid stream, no disc in drive e5 IO exception will be drawn.
Answer:
Convergent network
Explanation:
In networking, computers devices are connected together to communicate and share resources. Devices like routers and switches are examples of intermediate network devices and computers, smartphones, tablets and IP phones are all examples of end devices in a network.
In the past, dedicated networks are installed for separate voice, text and video packets. But as information technology evolves, one network is used for all three packets. This network is called convergent network.
Answer:
#include <iostream>
using namespace std;
int * reverse(int a[],int n)//function to reverse the array.
{
int i;
for(i=0;i<n/2;i++)
{
int temp=a[i];
a[i]=a[n-i-1];
a[n-i-1]=temp;
}
return a;//return pointer to the array.
}
int main() {
int array[50],* arr,N;//declaring three variables.
cin>>N;//taking input of size..
if(N>50||N<0)//if size greater than 50 or less than 0 then terminating the program..
return 0;
for(int i=0;i<N;i++)
{
cin>>array[i];//prompting array elements..
}
arr=reverse(array,N);//function call.
for(int i=0;i<N;i++)
cout<<arr[i]<<endl;//printing reversed array..
cout<<endl;
return 0;
}
Output:-
5
4 5 6 7 8
8
7
6
5
4
Explanation:
I have created a function reverse which reverses the array and returns pointer to an array.I have also considered edge cases where the function terminates if the value of the N(size) is greater than 50 or less than 0.
Answer:
29
Explanation:
for n=28:
--------------
Algorithm 1 performs f(n) = n2 + n/2 = 28*28 + 28/2 = 798
Algorithm 2 performs f(n) = 12*28 + 500 = 836
for n=29
--------------
Algorithm 1 performs f(n) = n2 + n/2 = 29*29 + 29/2 = 855.5
Algorithm 2 performs f(n) = 12*29 + 500 = 848
so, for n=29, algorithm 2 will be faster than algorithm 1
