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;
}
I think its B ir look on safari
Answer:
d. If X is NP - complete and Y is in NP then Y is NP - complete.
This can be inferred
Explanation:
The statement d can be inferred, rest of the statements cannot be inferred. The X is in NP complete and it reduces to Y. Y is in NP and then it is NP complete. The Y is not in NP complete as it cannot reduce to X. The statement d is inferred.
"C" is the correct answer, if its correct please mark me brainliest.
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