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

sinsun123

delete is used to release heap memory
So 1:
char * z = new char [6];
delete [] z;
can

So 2:
z = "abcde";
At this time, because "abcd" is no longer applying for heap memory,
delete [] z
error

Personal opinion, huh !!

zfy010317

cout << hex <<&z + k << endl;

Change to:
 cout << hex <<&z [k] << endl;

sun88888

zfy010317Brother:
cout<<hex<<&z+k<<endl;

Change to:
cout << hex <<&z[k] << endl;

I didn’t understand how to post spam,
I think at least you have to debug it to have a say
Otherwise, it is easy to mislead everyone

For the second question of the landlord, brotherch1074856is the positive solution
&z+k is an address that makes no sense at all
To get the addresses of all the elements in the array, just remove the address symbol&
Because z itself represents the first address of the array
cout<<hex<<z+k<<endl;

jjledu

Upstairs,zfy010317said that it was right, and it was you who was wrong (you said there was no debugging, did you debug)

ap2000ap

agree!

jjledu

Say Sorry tosun88888, you are also right, so is Space Hero:


    char z[]= "abcde";
    for(int k=0; k<5; k++)
    {
        cout<<hex<<(int*)(z+k)<<endl;//After testing, to output the address, the char* type address must be converted to int*, otherwise the characters will be output.
        cout<<hex<<(int*)&z[k]<<endl;//Space Hero
        cout<<z[k]<<endl;
    }

superandy

Landlord cout<<typeid(&z+k).name()<<endl;
cout<<typeid(*(z+k)).name()<<endl;
Just try and you will find out.

偶素菜鸟

Thank you so much.