Question about operator overloading and hard-coded strings.

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  1. #1
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    Question Question about operator overloading and hard-coded strings.

    Hello. I'm trying to review operator overloading in C++, and I've run into a problem that involves both it and typecasting. This line of code is giving me issues:

    Code:
    Integer integer = "34";
    The message is:

    Code:
    conversion from 'const char [3]' to non-scalar type 'Integer' requested|
    At first this was coming up with the following operator overload:

    Code:
    Integer Integer::operator=(const string &str)
    {
        value = str;
        return *this;
    }
    so I tried changing it to this to no avail:

    Code:
    Integer Integer::operator=(const char *str)
    {
        value = *str;
        return *this;
    }
    What's the deal? There's almost definitely a simple fix to this, but I'm at a loss at the moment as to what it is. Thanks!

  2. #2
    Registered User antred's Avatar
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    The problem is that, while your code looks like assignation, it really uses copy construction. You need to provide a copy constructor that takes a string literal.

    P.S. Also, operator = is supposed to return a REFERENCE not an actual copy.


    EDIT: As Elysia pointed out, this is not quite correct. It uses the constructor, not the copy constructor!
    Last edited by antred; 06-19-2012 at 12:29 PM. Reason: Because it's WRONG!

  3. #3
    C++まいる!Cをこわせ! Elysia's Avatar
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    Integer integer = "34";
    This is an initialization, the same as
    Integer integer("34");
    Initialization utilizes the constructor (not the copy constructor).

    Integer integer;
    integer = "34";
    This is assignment. It uses the assignment operator.
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  4. #4
    Registered User antred's Avatar
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    Quote Originally Posted by Elysia View Post
    Initialization utilizes the constructor (not the copy constructor).

    Meh, yeah you're right. Brainfart.

    @OP: Sorry about that.

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    Thanks. I'm kind of surprised it wasn't a more minor, technical detail about overloading syntax or typecasting.

  6. #6
    Algorithm Dissector iMalc's Avatar
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    You're probably better off making a constructor for it instead anyway rather than doing:
    Integer integer;
    integer = "34";
    Then thanks to the copy-constructor, it will allow you to do both.
    Of course you may also want to consider whether the constructor should be marked explicit.
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