将零赋给指针会破坏分配的内存吗？

Question

My question may seem trivial for some people, but I don't understand why this code:

double * a = new double [3];
a[0] = 1;
a[1] = 2;
a[2] = 3;

a = 0;

for(int i=0; i<3;i++)
cout << " a[" << i << "] = " << a[i] << endl;

delete [] a ; 

does not give the following result:

a[0] = 0;
a[1] = 0;
a[2] = 0;

Answer 1

The original memory allocated with new[] is leaked. You now have the equivalent of

double * a = 0; //or NULL

with memory leaks. Accessing the null pointer will give you undefined behavior. And no, assigning 0 to a pointer doesn't destroy the memory.

Answer 2

Your code will crash and/or produce some random result (if you assign some other constant to a ). Why?

In your code, a is a pointer to some region of the memory, which you are claiming for your own use (using new ). So you have a value of a like 0x05237484 (just a random example). So, you know that by that address 0x05237484 there are 24 bytes (3*8, 8 is the size of a double) bytes reserved for you. Essentially, when you say new double[5] , what you are saying to the runtime is: "now, find me a space enough to store an array of 5 doubles, and reserve it for me, so only I can use it". The runtime reserves memory for you and stores the address in the pointer you've provided - a

Then, what you do is overwriting the pointer a with some other value. This means that the memory is still reserved for you (because you didn't tell the runtime you no longer need it, by using delete or delete[] ), but you have forgotten about where it is stored.

Instead, your records now point to some other memory location - where there can be anything. So, either you will get some random values from where you are now pointing (if you happen to end up within same process memory), either your program will crash saying "Access violation" - meaning you've tried to access the memory belonging to another process/system/whatever. What you're doing here after assigning a = 0 is accessing 3 locations:

a[0] // which is a+0 == 0x00000000
a[1] // which is a+1 == 0x00000001
a[2] // which is a+2 == 0x00000002

You won't know what's at that location - but trying it, you'll either hit "access denied" error, or receive some random data (belonging to the other parts of application).

And, to make things worse, you've taken some memory to you but didn't release it - this means you've eaten up from your computer RAM for nothing. That's what called 'memory leak'

Answer 3

Assigning zero to a pointer doesn't destroy the allocated memory, that pointer stopes to point to that memory segment. And after assigning zero, delete[] will not delete that part of memory.