Answer:
cout << setprecision(2)<< fixed << number;
Explanation:
The above statement returns 12.35 as output
Though, the statement can be split to multiple statements; but the question requires the use of a cout statement.
The statement starts by setting precision to 2 using setprecision(2)
This is immediately followed by the fixed manipulator;
The essence of the fixed manipulator is to ensure that the number returns 2 digits after the decimal point;
Using only setprecision(2) in the cout statement will on return the 2 digits (12) before the decimal point.
The fixed manipulator is then followed by the variable to be printed.
See code snippet below
<em>#include <iostream> </em>
<em>#include <iomanip>
</em>
<em>using namespace std; </em>
<em>int main() </em>
<em>{ </em>
<em> // Initializing the double value</em>
<em> double number = 12.3456; </em>
<em> //Print result</em>
<em> cout << setprecision(2)<< fixed << number; </em>
<em> return 0; </em>
<em>} </em>
<em />
Answer:
A
Explanation:
The value will be the original stored in the int variables. This is because the method swap(int xp, int yp) accepts xp and yp parameters by value (They are passed by value and not by reference). Consider the implementation below:
public class Agbas {
public static void main(String[] args) {
int xp = 2;
int yp = xp;
swap(xp,yp); //will swap and print the same values
System.out.println(xp+", "+yp); // prints the original in values 2,2
int xp = 2;
int yp = 5;
swap(xp,yp); // will swap and print 5,2
System.out.println(xp+", "+yp); // prints the original in values 2,5
}
public static void swap(int xp, int yp){
int temp = xp;
xp = yp;
yp = temp;
System.out.println(xp+", "+yp);
}
}
