C++:将多态lambda传递给函数

这很好用:

template<class F>
bool func(F f)
{
    int a = 1, b = 2;
    float c = 3, d = 4;
    return f(a, b) || f(c, d);
}

int main()
{
    func([](auto a, auto b) -> bool { return a == b; });
    func([](auto a, auto b) -> bool { return a <= b; });
    func([](auto a, auto b) -> bool { return a < b; });
}

没有模板 auto 函数参数中的关键字没有意义。 要么将其作为模板,要么提供特定类型:

bool func(bool (* f)(int a, int b))
{
    int a = 1, b = 2;
    float c = 3, d = 4;
    return f(a, b) || f(c, d);
}

template<class T>
bool func2(T f)
{
    int a = 1, b = 2;
    float c = 3, d = 4;
    return f(a, b) || f(c, d);
}

…
func([](auto a, auto b) -> bool { return a <= b; });
func3([](auto a, auto b) -> bool { return a <= b; });

https://wandbox.org/permlink/cbit5S26a5pyrULf

好的,我找到了答案。更奇怪的是,我希望 func 函数也必须是这样的lambda函数:

auto func = [](auto f) -> bool
{
    int a = 1, b = 2;
    float c = 3, d = 4;
    return f(a, b) || f(c, d);
};