我使用了不能用 g++ 4.2.1 编译的 CRTP,可能是因为派生类本身就是一个模板?有谁知道为什么这不起作用,或者更好的是,如何使它起作用?示例代码和编译器错误如下。
资料来源:foo.C
#include <iostream>
using namespace std;
template<typename X, typename D> struct foo;
template<typename X> struct bar : foo<X,bar<X> >
{
X evaluate() { return static_cast<X>( 5.3 ); }
};
template<typename X> struct baz : foo<X,baz<X> >
{
X evaluate() { return static_cast<X>( "elk" ); }
};
template<typename X, typename D> struct foo : D
{
X operator() () { return static_cast<D*>(this)->evaluate(); }
};
template<typename X, typename D>
void print_foo( foo<X,D> xyzzx )
{
cout << "Foo is " << xyzzx() << "\n";
}
int main()
{
bar<double> br;
baz<const char*> bz;
print_foo( br );
print_foo( bz );
return 0;
}
编译器错误
foo.C: In instantiation of ‘foo<double, bar<double> >’:
foo.C:8: instantiated from ‘bar<double>’
foo.C:30: instantiated from here
foo.C:18: error: invalid use of incomplete type ‘struct bar<double>’
foo.C:8: error: declaration of ‘struct bar<double>’
foo.C: In instantiation of ‘foo<const char*, baz<const char*> >’:
foo.C:13: instantiated from ‘baz<const char*>’
foo.C:31: instantiated from here
foo.C:18: error: invalid use of incomplete type ‘struct baz<const char*>’
foo.C:13: error: declaration of ‘struct baz<const char*>’