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Author: 偶素菜鸟

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

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Post time: 2020-5-23 17:15:01
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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 !!
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Post time: 2020-5-29 10:15:01
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cout << hex <<&z + k << endl;

Change to:
 cout << hex <<&z [k] << endl;
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Post time: 2020-6-10 16:30:01
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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;
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Post time: 2020-6-17 10:00:01
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Upstairs,zfy010317said that it was right, and it was you who was wrong (you said there was no debugging, did you debug)
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Post time: 2020-6-18 10:30:01
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agree!
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Post time: 2020-6-18 20:15:01
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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;
    }
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Post time: 2020-6-19 17:15:02
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Landlord cout<<typeid(&z+k).name()<<endl;
cout<<typeid(*(z+k)).name()<<endl;
Just try and you will find out.
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 Author| Post time: 2020-7-17 13:45:01
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Thank you so much.
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