Can anyone explain how he each of the answer and also give me a clear answer for each questions also . Thanks

P.S I do not have the full program.

#include using namespace std; int main( ) { typedef int my_2darray[1][1]; my_2darray b[3][2]; cout

It would be something similar to this program.

W the next line prints 0012FFAC coute The address of bisp address of b+1 is: "b+lecendl;4 coutse The address of &b is:bocendl; Ws coutss The address of &b is:&blecendl endl; 6 return 0 a. At point Ithe program outputs 210. Why? ( points) b. At polnt 2 the program outputs 4. Why? (2 points) c. At point 3 the program outputs 105. Why? 3 points) having a he d, what will be printed at point 42001300 ,r HoS. S2- l2 (1+3poins 0012 FFAC 1219 21 lbt 0 01300S e. What will be printed at point 5? (I+ 3 points) g print 0012AC/ because Ibis kesmp fue acecho, f b opva f What will be printed at point 6? (+3 points) 2. l0 2 013007e 23 b