Question

我正在尝试将绑定成员函数传递给例程，并且结果类型由模板类确定：

template< class Fun, class P>
auto splat(double start, double finish, int steps,  Fun fun, double bias)
{
    typedef BOOST_TYPEOF(Fun) Fun_type;
    typedef boost::function_traits<Fun_type> function_traits;
    typedef boost::function_traits<Fun_type>::result_type P;
    vector<P> out = vector<P>(steps);
    double a = steps;
    a = a / 2;
    double step = (finish - start) / (steps - 1);
    double param = start;
    for (int i = 0; i < steps; i++)
    {
        out[i] = fun(param);
        param += step*(1 + accel*(i - a - 1) / (a - 1));
    }
    return out;
}

调用顺序是：

std::function<BallLib::Point(double)> pFun = std::bind(&BallPath::path, one, _1);
Point ppp = pFun(0.0);
vector<Point> line = testspread::splat(0.0, 3.1415, 10, pFun, 0.0);

无法编译严重性代码描述项目文件行错误C2783'std :: vector＆gt; testspread :: splat（double，double，int，Fun，double）'：无法推断'P'的模板参数

如何确定P？

Answer 1

在你的签名

template <class Fun, class P>
auto splat(double start, double finish, int steps, Fun fun, double bias);

您需要（不可扣除）类型P（您不能使用）。

将您的代码更改为

template <class Fun>
auto splat(double start, double finish, int steps, Fun fun, double bias)
{
    typedef BOOST_TYPEOF(Fun) Fun_type;
    typedef boost::function_traits<Fun_type> function_traits;
    typedef boost::function_traits<Fun_type>::result_type P;
    vector<P> out(steps);
    const double a = steps / 2;
    const double step = (finish - start) / (steps - 1);
    double param = start;
    for (int i = 0; i < steps; i++)
    {
        out[i] = fun(param);
        param += step*(1 + accel*(i - a - 1) / (a - 1));
    }
    return out;
}

应该解决您的问题。

传递绑定成员函数

1 个答案: