Question

Can anyone help me, why I’m getting an error message while trying to free the allocated memory: Heap corruption detected. CTR detected the application wrote the memory after end of heap buffer.

char *ff (char *s){
    char *s1 = new char [strlen(s)];
    strcpy(s1, s);
    return s1;
}

int _tmain(int argc, _TCHAR* argv[])
{
    char *s = new char [5];

    strcpy(s, "hello");
    char *s2 = ff(s);

    delete []s;     // This works normal
    delete []s2;    // But I get an error on that line
    return 0;
}

Answer 1

char *s = new char [5];
strcpy(s, "hello");

Causes Undefined behavior(UB).
You are writing beyond the bounds of allocated memery. You allocated enough memory for 5 characters but your string has 6 characters including the \0.

Once your program has caused this UB, all bets are off and any behavior is possible.

You need:

char *s = new char [strlen("hello") + 1];

In fact the ideal solution is to use std::string and not char *. These are precisley the mistakes which std::string avoids. And there is no real need of using char * instead of std::string in your example.
With std::string:

• You don’t need to new anything
• You don’t need to delete anything &
• You can do everything with std::string, that you do with char *.