将派生类传递给专门用于基类的模板函数 [英] Passing a derived class to a template function specialized with base class

问题描述

以下代码可正常编译，但会产生链接器错误:

The following code compiles fine, but produces a linker error:

class Base {};

class Derived : public Base {};

template <typename T>
void f(const T& value);

template <>
void f(const Base& value) {
    // ...
}

int main() {
    Base b;
    f(b);

    Derived d;
    f(d); // This line causes linker error.

    return 0;
}

是否可以在不为派生类添加重复的 f()专业化条件的情况下，使该代码进行编译和链接?

Is it possible to make this code compile and link without adding a duplicate f() specialization for the derived class?

谢谢.

P.S我正在使用clang，Apple LLVM 6.0版(clang-600.0.56)(基于LLVM 3.5svn).

P.S I'm using clang, Apple LLVM version 6.0 (clang-600.0.56) (based on LLVM 3.5svn).

P.P.S链接器错误是:

P.P.S The linker error is:

Undefined symbols for architecture x86_64:
  "void f<Derived>(Derived const&)", referenced from:
      _main in test2-4245dc.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

推荐答案

发生链接器错误是因为您尚未定义(仅声明)函数模板

The linker error occurs because you haven't defined, only declared, the function template

template <typename T>
void f(const T& value);

为避免这种情况，请定义上面的模板，您的代码将被编译，但是大概它仍然无法满足您的要求.

To avoid it, define the above template and your code will compile, but presumably it still doesn't do what you want.

当使用类型为 Derived 的参数调用 f 时，与您的专业化相比，上面的模板更好地匹配，因为后者需要从派生到基数的转换，而前者没有.

When calling f with an argument of type Derived the above template is a better match compared to your specialization because the latter requires a derived-to-base conversion while the former doesn't.

您可以使用 enable_if 仅在推导的模板参数类型不是 Base 或从 Base 派生的类型时，才允许第一个模板参与重载解析.

You can achieve the behavior you want by using enable_if to allow the first template to participate in overload resolution only if the deduced template argument type is not Base or a type derived from Base.

template <typename T>
typename std::enable_if<!std::is_base_of<Base, T>::value>::type
    f(const T& value) {

}

并更改另一个 f ，所以它不是专门化的，而只是重载.

And change the other f so it's not a specialization, but just an overload.

void f(const Base& value) {
    // ...
}

实时演示

通常，重载函数模板胜于专业化.阅读此，以了解功能模板专业化的陷阱.

In general, prefer overloading function templates over specialization. Read this to learn about the pitfalls of function template specialization.

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