C++ 'std::out_of_range' error when turning string into character array

Question

So, basically I'm creating a program that at certain points needs to display text one character at a time with intervals between each character. I created a function that I can pass a string into that is supposed to display the string one character at a time slowly. The only problem is when I'm taking each character from the string, I get the error -> "terminate called after throwing an instance of 'std::out_of_range'what(): basic_string::at"

I've been trying to find the problem for quite a while and all the other code seems to be working, but code that puts the character from the string into the array of characters I made, and I have no idea how to fix it. Any help is much appreciated.

Code:

std::string SlowText(std::string s)
{
    int L = s.length();
    char *Text;
    Text = new char[L];
    int c = L;
    while(c > 0)
    {
        Text[c] = s.at(c);
        --c;
    }
    c = L;
    while(c > 0)
    {
        std::cout << Text[c];
        Sleep(250);
        --c;
    }
return "";
}

Answer 1

The reason is that L is the length of the array and you do this:

c = L;

because of 0 indexing you are starting past the end of the string. Try this:

c = L-1;

Of course since this is c++, I am going to tell you the standard thing which is don't use arrays! your code could be this:

std::string SlowText(std::string s)
{
    for (auto b = s.rend(), e = s.rbegin(); b != e; b++)
    {
         std::cout << *b << std::flush;
         sleep(250)
    }
    return ""; //Also why have this?
}