Doubt in my code

#1

What is line 5, 6, 7 telling us ?

#2

hello @yashsharma4304

  int* ptr = new int[100];
    cout<<&ptr<<endl;            
    //&ptr will give address of ptr variable    
    cout<<*ptr<<endl;
// since ptr is having base address of array, *ptr will give u value present at that address (at index 0)  which is grabage.
    cout<<ptr<<endl;
// this  will print base address of array becuase ptr is holding base address of array
#3

Ok so if I write this line after assigning the values to dynamic array then it will print 0. But why the base address of dynamic array is changing everytime ? Everytime I run the program does the dynamic array
allocated with new memory ?

#4

yeah the address that u r getting may or may not be same.

it will depend on what all memory blocks are free in ur system accordingly os will allocate memory at appropriate location and give its base address

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#5

Ok thank you very much for resolving my doubt :slightly_smiling_face:

#6

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