Answer:
Used the command syntax; awk -F":" '{ print "username: " $<number location> "\t\tuid:" $<number location> }' <target folder>
Explanation:
Linux operating system is a fast open-source computer platform for programmers and network administrators. Its system is arranged in a hierarchical tree structure with the root represented as "/" (for absolute path).
The passwd is a folder in the Linux OS that holds the login details of all users in the system network. The 'awk' is one of the commands used to get information from a file in a folder. It prints out the result by specifying the location of the values (like the username and user id) as a variable (with prefix '$') and then the target folder.
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
Give the type and value of each result of the following Java expressions. a. (5 / 2) * 2.0 type is float value is 4.0 b. (5/2.0) * 2 type is integer value is 5c. "1.3" + "5.2" type is string value is 1.35.2d. 1 + 7.0 + "2" + "x" this will produce an error because we cannot add variables of different type
Answer: static
Explanation:
variables when declared static gets called statically meaning whenever a function call is made it get stored and it is not required to get the variable again when the function is again called. There scope is beyond the function block
