Question

我有一个模板类：

template <typename T> struct randimpl {
    // default is to not compile, have to pass T to something to get assert
    // evaluated during second phase expansion, is_void is an arbitrary choice.
    static_assert(
        std::is_void<T>::value && false, "random() not implemented for type"
    );
};

这只是random()函数的代理，可以为用户定义的类型重载：

// random function, proxies request through randimpl
template <typename T> T random() {
    return randimpl<T>::random();
}

我试图对_point1d<T>类型进行部分专业化处理：

// generate random instances _point1d<T>
template <typename T>
struct randimpl<_point1d<T>> {
    static _point1d<T> random() {
        return _point1d<T>(random<T>());
    }
};

据我所知，它是合适的语法，但是当我编译时，我得到：

inc/geometry.h: In static member function ‘static _point1d<T> randimpl<_point1d<T> >::random()’:
inc/geometry.h:287:31: error: expected primary-expression before ‘(’ token
             return _point1d<T>(random<T>());
                               ^
inc/geometry.h:287:40: error: expected primary-expression before ‘>’ token
             return _point1d<T>(random<T>());
                                        ^
inc/geometry.h:287:42: error: expected primary-expression before ‘)’ token
             return _point1d<T>(random<T>());

我的语法是否关闭？这似乎应该有效，但我不明白为什么它不接受它。

Answer 1

问题是static成员函数的名称random隐藏了全局函数，它试图调用自身（即递归调用）。但它不是一个模板，与它的调用方式不符。

您可以添加:: (scope resolution)来指定全局random，例如

// generate random instances _point1d<T>
template <typename T>
struct randimpl<_point1d<T>> {
    static _point1d<T> random() {
        return _point1d<T>(::random<T>());
        //                 ^^
    }
};

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部分专门化c ++中的模板类

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