Usually back to their start point. Or a random place on the map.
Answer:
10
Explanation:
num1=3 and num2=2
num3=myproc(num1)+myproc(num2)
myproc(num1) results 6
myproc(num2) results 4
when we add both e get 10
Answer:
public class print{
public static void fillArray(int[] arr, int initialValue){
int n = arr.length;
for(int i=0;i<n;i++){
arr[i] = initialValue++;
}
for(int i=0;i<n;i++){
System.out.print(arr[i]+" ");
}
}
public static void main(String []args){
int[] array = new int[5];
int initialValue =3;
fillArray(array,initialValue);
}
}
Explanation:
Create the function with two parameter first is array and second is integer.
Then, declare the variable and store the size of array.
Take the for and fill the array from the incremented value of initialValue by 1 at every run of loop.
After loop, print the element of the array.
Create the main function which is used for calling the function and also declare the array with size 5 and initialValue with 3. After that, call the function with this argument.
The cursor should move right but if not give me more information to correct my statement
Answer:
a) AL will contains 0011 1100
Explanation:
In assembly language, shifting bits in registers is a common and important practice. One of the shifting operations is the SHR AL, x where the x specifies that the bits be shifted to the right by x places.
SHR AL, 2 therefore means that the bits contained in the AL should be shifted to the right by two (2) places.
For example, if the AL contains binary 1000 1111, the SHR AL, 2 operation will cause the following to happen
Original bit => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) | (0) |
Notice;
(i) that there are two shifts - one at a time.
(ii) that the bits in bold face are the bits in the AL after the shift. Those that in regular face are those in the carry flag.
(iii) that the new bits added to the AL after a shift are the ones in bracket. They are always set to 0.
