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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T16:26:39+00:00 2026-05-14T16:26:39+00:00

I have the following code, which uses a Unicode string class from a library

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I have the following code, which uses a Unicode string class from a library that I’m writing:

#include <cstdio>
#include "ucpp"
main() {
  ustring a = "test";
  ustring b = "ing";
  ustring c = "- -";
  ustring d;
  d = "cafe\xcc\x81";
  printf("%s\n", (a + b + c[1] + d).encode());
}

The encode method of the ustring class instances converts the internal Unicode into a UTF-8 char *. However, because I don’t have access to the char class definition, I am unsure on how I can define an implicit typecast (so that I don’t have to manually call encode when using with printf, etc).

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  1. Editorial Team
    Editorial Team
    2026-05-14T16:26:40+00:00Added an answer on May 14, 2026 at 4:26 pm

    First, I would recommend that you consider not providing an implicit conversion. You may find that the situations where unexpected conversions are not caught as errors outweighs the cost of calling encode when you want a char*.

    If you do decide to provide an implicit conversion you declare it like this (inside your class definition.

    operator char*();
    

    You might be able to make the method const, in which case you can use:

    operator char*() const;
    

    Usually you would also want to return a pointer to a non-modifiable buffer:

    operator const char*() const;
    

    In the body of your function you should return an appropriate pointer. As an implicit conversion clients wouldn’t expect to have to free the returned buffer so if you need to make a special buffer for your return value you will have to maintain a pointer to this buffer until a suitable point to free it. Typically such a suitable moment might be the next mutating operation on your class object.

    Note that as printf takes any number and type of optional arguments you would still need to cast your class object in any case.

    printf("%s\n", static_cast<const char*>(a + b + c[1] + d));
    

    or

    printf("%s\n", (const char*)(a + b + c[1] + d));
    

    Both of these are more verbose than an explicit call to encode.

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