可变参数模板的“呼叫无匹配功能”

Question

I have a class that is designed to dynamically load a .dll or .so or equivalent. From there, it will return pointers to whatever function you're trying to find. Unfortunately, I've come across two issues in my implementation.

1. If I use the 'dumb' function returning void* as pointers to functions, I get warning: ISO C++ forbids casting between pointer-to-function and pointer-to-object when I try to manipulate them into a form I can use.
2. If I try using the 'smart' function with variadic templates and type safety, I can't get it to compile. error: no matching function for call to 'Library::findFunction(std::string&)' is the only thing that awaits me here. As you can see from the code below, this should match the function signature. Once it does compile, issue 1 will be present here too.

For reference, I am compiling under Ubuntu 10.10 x86_64 with g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5 . I have also tried compiling with g++-4.5 (Ubuntu/Linaro 4.5.1-7ubuntu2) 4.5.1 however this does not change anything.

#include <string>
#include <stdio.h>

class Library
{
public:
    Library(const std::string& path) {}
    ~Library() {}

    void* findFunction(const std::string& funcName) const
    {
        // dlsym will return a void* as pointer-to-function here.
        return 0;
    }

    template<typename RetType, typename... T>
    RetType (*findFunction(const std::string& funcName))(T... Ts) const
    {
        return (RetType (*)(...))findFunction(funcName);
    }

};

int main()
{
    Library test("/usr/lib/libsqlite3.so");

    std::string name = "sqlite3_libversion";
    const char* (*whatwhat)() = test.findFunction<const char*, void>(name);
    // this SHOULD work. it's the right type now! >=[

    //const char* ver3 = whatwhat();
    //printf("blah says \"%s\"\n", ver3);
}

Answer 1

I think you need to change the returned function signature to T... instead of just ... otherwise it expects a variable arglist even when it should just be empty.

template<typename RetType, typename... T>
RetType (*findFunction(const std::string& funcName))(T... Ts) const
{
    return (RetType (*)(T...))findFunction(funcName);
}

Then call it without the void in the type list and it should work:

const char* (*whatwhat)() = test.findFunction<const char*>(name);

With these changes it compiles for me on gcc-4.5.1 : http://ideone.com/UFbut