I do n’t understand a question about enrollment, everyone helps

偶素菜鸟

I wrote a program like this

char * z = new char [6];
z = "abcde";
for (int k = 0; k <5; k ++)
{
cout << hex <<&z + k << endl;
cout << * (z + k) << endl;
}

delete [] z;

The previous operation is quite normal, and then delete [] z; there, it can not run, what is the reason? I guess it has something to do with that const, please help.

Then, in order to run this program, I wrote this:

char z [] = {'a', 'b', 'c', 'd', 'e'};
Ranch
          for (int k = 0; k <5; k ++)
{
cout << hex <<&z + k << endl;
cout << * (z + k) << endl;
}

No need to delete, there is nothing wrong, but I suddenly found a strange phenomenon. This is its output:
0012FF24
a
0012FF29
b
0012FF2E
c
0012FF33
d
0012FF38
e
How can their memory locations differ by five? My original experiments were all four bits apart. Isn't a computer word four bytes long? Why is that?

偶素菜鸟

Everyone helps, C pointer is really annoying.

csn_killer

char * z = new char [6];
z = "abcde"; // pointed elsewhere ... you didn't use your dynamically allocated memory to save "abcde"
                      // strcpy (z, "abcde"); // That's fine
for (int k = 0; k <5; k ++)
{
cout << hex <<&z + k << endl;
cout << * (z + k) << endl;
}

delete [] z;

yexilove

z = "abcde";

Change the pointer of z, and of course an error will occur when delete.

strcpy (z, "abcde");

diamond52

char * z = new char [6];
========================
This sentence your z points to the memory allocated on the heap

z = "abcde";
======
In this sentence you point z elsewhere

delete [] z;
======
This sentence you want to release the memory on the heap, but z does not point to the heap memory, so

diamond52

Solution
Save a pointer to the memory on the heap
For release
char * z = new char [6];
char * p = z; // here
z = "abcde";
// ...
delete [] p; // ok

diamond52

Personal opinions, experts advise

vam811029

second question........
cout << hex <<&z + k << endl; The output is an address other than the array z.

ch1074856

for (int k = 0; k <5; k ++)
{
cout << hex <<&z + k << endl; //&z points to the address of a data structure whose type is char [5]. The value of&z + k is actually equal to
                       // address of z + sizeof (char [5]) * k
cout << * (z + k) << endl;
}

偶素菜鸟

How can I get the addresses of all the elements in an array?