difference between engineering and architecture are.
A engineer is a person whose job involves designing and building engines, machines, roads, bridges ,etc .
While architect design buildings only.
Answer:
To see who reads the file, open “Windows Event Viewer”, and navigate to “Windows Logs” → “Security”. There is a “Filter Current Log” option in the right pane to find the relevant events. If anyone opens the file, event ID 4656 and 4663 will be logged.
Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
Answer:
#include<stdio.h>
//declare a named constant
#define MAX 50
int main()
{
//declare the array
int a[MAX],i;
//for loop to access the elements from user
for(i=0;i<MAX;i++)
{
printf("\n Enter a number to a[%d]",i+1);
scanf("%d",&a[i]);
}
//display the input elements
printf("\n The array elements are :");
for(i=0;i<=MAX;i++)
printf(" %d ",a[i]);
}
Explanation:
PSEUDOCODE INPUTARRAY(A[MAX])
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT “Enter a number to A[I]”
READ A[I]
[END OF LOOP]
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT A[I]
[END OF LOOP]
RETURN
ALGORITHM
ALGORITHM PRINTARRAY(A[MAX])
REPEAT FOR I<=1 TO MAX
PRINT “Enter a number”
INPUT A[I]
[END OF LOOP]
REPEAT FOR I<=1 TO MAX
PRINT A[I]
[END OF LOOP]
