C ++指针组成一个数组

Question

I am new to C++ ... I am playing with pointers ... This code uses a char ptr as an array .

#‎include‬<iostream>
using namespace std;

int main (){

    cout << "Playground "<<endl;

    const short max=10;

    char * str=new char ;

    for (short i=0;i<max;i++)
        *(str+i)=char(i+100);

    for (short i=0;i<max;i++)
        cout <<str+i<< char(3) <<endl;

    for(short i=0;i<max;i++)
        delete (str+i);

    return 0;
}

---

But I am not sure if delete (str+i) works or not , and why ? But I think it does not work because I have run the program many times and although the string is printed :

Playground
defghijklm
efghijklm
fghijklm
ghijklm
hijklm
ijklm
jklm
klm
lm
m

I have this error message

try #0
*** Error in `./p': free(): invalid pointer: 0x09d90009 ***
Aborted (core dumped)

try #1
*** Error in `./p': free(): invalid pointer: 0x08453009 ***
Aborted (core dumped)

try #2
*** Error in `./p': free(): invalid pointer: 0x0863c009 ***
Aborted (core dumped)

etc ...

Because the invalid pointer keep changing , I have the feeling that the objects has not been deleted or garbage collected and every time I run the code I use another area of the memory ...

Finally I am starting to get why regular array are better ...

*from the beginning a contiguous space in memory is reserved for the data

*not need to worry about delete ...

Those are just my guesses...

Answer 1

You have out of bound access:

char* str = new char; // single element

should be

char* str=new char[max]; // array of char

and you have to release memory with

delete [] str;

but to avoid to manage memory manually, you may use std::string or std::vector<char>

Answer 2

You are only making one dynamic memory allocation, and for just 1 char, but you are deleting dynamically allocated memory multiple times in the last for-loop - you can only do one delete for every new.

Additionally, in the first for-loop you assign new char values to memory outside of what you allocated, which means you are overwriting memory that isn't yours.