具有模板参数而非函数参数的Lambda函数 [英] Lambda functions with template parameters, not in function parameters

问题描述

为什么第一次调用无法编译？

Why does the first call not compile?

auto get1 = []<int B>() { return B; };
auto get2 = []<typename B>(B b) { return b; };

int main()
{
    get1<5>(); // error: no match for operator<
    get2(5);   // ok
}

我之所以使用它，是因为它在代码。

The reason I use this, is an expression repeated many times in code.

我当然可以使用真实的函数模板，但是我很好奇。

Of course I can use a real function template, but just I am curious WHY.

推荐答案

如果考虑与 get1 等效的类类型，这将更容易理解：

This is easier to understand if you consider what the equivalent class type looks like to your get1:

struct get1_t {
    template <int B> operator()() const { return B; }
};

get1_t get1;

get1<5>(); // error

您试图为调用运算符提供一个显式的模板参数，但是从语法上讲，正在做为 get1 本身提供模板参数的操作（即，好像 get1 是一个变量模板）。为了为调用运算符提供模板参数，您必须直接执行以下操作：

You're trying to provide an explicit template parameter to the call operator, but syntactically you're doing what looks like providing template parameters for get1 itself (i.e. as if get1 were a variable template). In order to provide the template parameter for the call operator, you have to do that directly:

get1.operator()<5>(); // ok

或重组呼叫运算符以采用可推论的东西：

Or restructure the call operator to take something deducible:

template <int B> struct constant { };
get1(constant<5>{});

---

或者将整个结构重组为实际的看起来像这样的变量模板：

---

Or restructure the whole thing to actually be the variable template that it looks like it is:

template <int B>
auto get1 = [] { return B; };

现在， get1< 5> 本身您正在调用的lambda。也就是说，除了具有调用运算符模板的lambda之外，我们还有一个变量模板lambda本身不是模板。

Now, get1<5> is itself a lambda, that you're invoking. That is, rather than a lambda with a call operator template we have a variable template lambda that is itself not a template.

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