Answer:
hope this helps. I am also a learner like you. Please cross check my explanation.
Explanation:
#include
#include
using namespace std;
int main()
{
int a[ ] = {0, 0, 0}; //array declared initializing a0=0, a1=0, a3=0
int* p = &a[1]; //pointer p is initialized it will be holding the address of a1 which means when p will be called it will point to whatever is present at the address a1, right now it hold 0.
int* q = &a[0]; //pointer q is initialized it will be holding the address of a0 which means when q will be called it will point to whatever is present at the address a0, right now it hold 0.
q=p; // now q is also pointing towards what p is pointing both holds the same address that is &a[1]
*q=1
; //&a[0] gets overwritten and now pointer q has integer 1......i am not sure abut this one
p = a; //p is now holding address of complete array a
*p=1; // a gets overwritten and now pointer q has integer 1......i am not sure abut this one
int*& r = p; //not sure
int** s = &q; s is a double pointer means it has more capacity of storage than single pointer and is now holding address of q
r = *s + 1; //not sure
s= &r; //explained above
**s = 1; //explained above
return 0;
}
Answer:
A real-time operating system is an operating system designed to support real-time applications that, usually without buffer delays, process data as it comes in. A real-time system is a time-bound system that has fixed, well defined time constraints.
Answer:
B. Doesn’t get you to your location faster.
Explanation:
Following the car in front of you at a closer distance doesn’t get you to your location faster.
Answer:
(a) What is the best case time complexity of the algorithm (assuming n > 1)?
Answer: O(1)
(b) What is the worst case time complexity of the algorithm?
Answer: O(n^4)
Explanation:
(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1)
.
(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).
