Why can't it recognize template method?

This is a discussion on Why can't it recognize template method? within the C++ Programming forums, part of the General Programming Boards category; When I try this code, it says "no matching function for..." Code: template <typename MyType> std::vector<MYType> get( ...elided... ) { ...

  1. #1
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    Why can't it recognize template method?

    When I try this code, it says "no matching function for..."

    Code:
    template <typename MyType>
    std::vector<MYType> get( ...elided... )
    {
         ...elided...
    }
    
    int main()
    {
        std::vector<std::vector<std::string> ret = get( ...elided... );
    }

  2. #2
    Master Apprentice phantomotap's Avatar
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    Presumably the problem is in the "...elided..." bits. The compiler can only "guess" a type from a template that it gets for a parameter.

    Soma

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    Quote Originally Posted by phantomotap View Post
    Presumably the problem is in the "...elided..." bits. The compiler can only "guess" a type from a template that it gets for a parameter.

    Soma
    No, I believe it's because the only possible differentiation in the instances (specializations) of the function is in the return type. So I think (and I could be wrong) that overloaded methods have to differ by args not just return type. Is that correct?

  4. #4
    Master Apprentice phantomotap's Avatar
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    The compiler can only "guess" a type from a template that it gets for a parameter.
    O_o

    Well, obviously that's not true. I don't even know why I said it.

    For a free function used as a parameter to a constructor, this context, it is true.

    Soma
    Last edited by phantomotap; 04-11-2008 at 08:58 PM.

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    Quote Originally Posted by phantomotap View Post
    O_o

    Well, obviously that's new true. I don't even know why I said it.

    For a free function used as a parameter to a constructor, this context, it is true.

    Soma
    Whoops. Sorry, I misread your comment.

    Thanks!

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    C++まいる!Cをこわせ! Elysia's Avatar
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    Quote Originally Posted by 6tr6tr View Post
    No, I believe it's because the only possible differentiation in the instances (specializations) of the function is in the return type. So I think (and I could be wrong) that overloaded methods have to differ by args not just return type. Is that correct?
    Yes, this is true. And overloaded function cannot merely differ by return type. The arguments must differ, as well.
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    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

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    Algorithm Dissector iMalc's Avatar
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    As Elysia said:
    ...elided... as well.
    My homepage
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    Cat without Hat CornedBee's Avatar
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    Quote Originally Posted by Elysia View Post
    Yes, this is true. And overloaded function cannot merely differ by return type. The arguments must differ, as well.
    While this is correct, it is irrelevant to the problem. Template instantiations do not overload each other. fn<Foo>() and fn<Bar>() are completely different functions. This is how, for example, lexical_cast, any_cast, and the many variants of get() work in Boost:
    Code:
    template <typename T>
    T any_cast(const any &a);
    
    // Usage:
    any a(100);
    int i = any_cast<int>(a);
    
    template <int N, typename T1, typename T2, COMPLICATED_PP_METAPROGRAM>
    complicated_metaprogram get(const tuple<T1, T2, COMPLICATED_PP_METAPROGRAM> &t);
    
    // Usage:
    tuple<int, float, Foo> tpl;
    int i = get<0>(tpl);
    float f = get<1>(tpl);
    Foo foo = get<2>(tpl);
    Note that in both cases, the arguments to the variants are always the same. Template arguments are sufficient to distinguish functions.
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    Quote Originally Posted by CornedBee View Post
    While this is correct, it is irrelevant to the problem. Template instantiations do not overload each other. fn<Foo>() and fn<Bar>() are completely different functions. This is how, for example, lexical_cast, any_cast, and the many variants of get() work in Boost:
    Code:
    template <typename T>
    T any_cast(const any &a);
    
    // Usage:
    any a(100);
    int i = any_cast<int>(a);
    
    template <int N, typename T1, typename T2, COMPLICATED_PP_METAPROGRAM>
    complicated_metaprogram get(const tuple<T1, T2, COMPLICATED_PP_METAPROGRAM> &t);
    
    // Usage:
    tuple<int, float, Foo> tpl;
    int i = get<0>(tpl);
    float f = get<1>(tpl);
    Foo foo = get<2>(tpl);
    Note that in both cases, the arguments to the variants are always the same. Template arguments are sufficient to distinguish functions.
    You're right!

    The problem was in what I was specializing to. I'd
    accidentally done:

    Code:
         vector<vector<string> > ret = get<vector<string> >( c );
    when it should have been:

    Code:
         vector<vector<string> > ret = get<string>( c );
    Thanks!

  10. #10
    Master Apprentice phantomotap's Avatar
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    *grumble*

    My second edit was not accepted...

    Anyway, In rare cases the compiler will choose a function based only on the return type. With the right bit of machinery you can force the compiler to examine the target result. If it wasn't for the silly rules regarding when precisely a type is known, even this would not be necessary--explicitly stating the template arguments.

    (I thought the OP was asking about how to do that in the case of a constructor.)

    Also, are you, the OP, going to at least tell the DEVX forum that you found a solution?

    Soma

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