为什么这段代码不编译？

Question

我有这样的情况:

struct Foo
{
    void Barry() { }
};

struct Bar : private Foo
{
    template <class F> void Bleh(F Func) { Func(); }
};

struct Fooey : public Bar
{
    void Blah() { Foo f; Bar::Bleh(std::bind(&Foo::Barry, &f)); }
};

它不编译(g ++ 4.7.3).有错误:

test.cpp: In member function ‘void Fooey::Blah()’:
test.cpp:4:1: error: ‘struct Foo Foo::Foo’ is inaccessible
test.cpp:15:23: error: within this context
test.cpp:4:1: error: ‘struct Foo Foo::Foo’ is inaccessible
test.cpp:15:47: error: within this context

但是,如果我这样做:

class Fooey;
void DoStuff(Fooey* pThis);

struct Fooey : public Bar
{
    void Blah() { DoStuff(this); }
};

void DoStuff(Fooey* pThis)
{
    Foo f;
    pThis->Bleh(std::bind(&Foo::Barry, &f));
}

它编译得很好.这背后的逻辑是什么？

Answer 1

这里

struct Fooey : public Bar
{
    void Blah() { Foo f; Bar::Bleh(std::bind(&Foo::Barry, &f)); }
};

name lookup for Foo查找Bar由于Bar私有继承而无法访问的基类.

要修复它,请完全限定名称:

    void Blah() { ::Foo f; Bar::Bleh(std::bind(&::Foo::Barry, &f)); }