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I have created a new string type with a memory pool allocator. Lets
presume the allocator works, my type is defined as:


typedef std::basic_string<char, std::char_traits<char>,
pool_allocator<char> > NewStr;

This is the same as std::string except with a different allocator,
right?

I want to be able to implicitly convert from std::string to NewStr
like:

std::string a("hi");
NewStr b("there");
a = b;

or
b = a;

which doesnt work, becasue they are different types. So my question
is, if i wanted to do this:

class MyString : public NewStr
{ public:
MyString &operator=(const std::string &other)
{
// what is the best way to do this?
*this = other.c_str(); /// doesnt look good...
return *this;
}

of course the same question could apply to the copyCtor too...

and, to go the other way, should i do the same thing for the

operator const std::string&() {...} method?

Is there a better way to create my type?

Jim

Jim Kogler wrote:
> I have created a new string type with a memory pool allocator. Lets
> presume the allocator works, my type is defined as:
>
>
> typedef std::basic_string<char, std::char_traits<char>,
> pool_allocator<char> > NewStr;
>
> This is the same as std::string except with a different allocator,
> right?

Right. It makes it a totally different, unrelated type, nonetheless.

>
> I want to be able to implicitly convert from std::string to NewStr
> like:
>
> std::string a("hi");
> NewStr b("there");
> a = b;
>
> or
> b = a;
>
> which doesnt work, becasue they are different types. So my question
> is, if i wanted to do this:
>
> class MyString : public NewStr
> { public:
> MyString &operator=(const std::string &other)
> {
> // what is the best way to do this?
> *this = other.c_str(); /// doesnt look good...

Probably something like

this->assign(other.begin(), other.end());

> return *this;
> }
>
> of course the same question could apply to the copyCtor too...

You may use 'assign' there too.

>
> and, to go the other way, should i do the same thing for the
>
> operator const std::string&() {...} method?

No. What would it be a reference to? You can only return an object
from that, and not a reference:

operator std::string() const {
return std::string(this->begin(), this->end());
}

>
> Is there a better way to create my type?

Not if you want your implicit conversions.

Victor