如何使用lambda表达式作为模板参数？ [英] How to use a lambda expression as a template parameter?

问题描述

如何使用lambda表达式作为模板参数？例如。作为初始化std :: set的比较类。

How to use lambda expression as a template parameter? E.g. as a comparison class initializing a std::set.

以下解决方案应该可以工作，因为lambda表达式只创建一个匿名结构，它应该适合作为模板参数。

The following solution should work, as lambda expression merely creates an anonymous struct, which should be appropriate as a template parameter. However, a lot of errors are spawned.

代码示例：

struct A {int x; int y;};
std::set <A, [](const A lhs, const A &rhs) ->bool {
    return lhs.x < rhs.x;
    } > SetOfA;

错误输出（我使用g ++ 4.5.1编译器和--std = c ++ 0x编译标志）：

Error output (I am using g++ 4.5.1 compiler and --std=c++0x compilation flag):

error: ‘lhs’ cannot appear in a constant-expression
error: ‘.’ cannot appear in a constant-expression
error: ‘rhs’ cannot appear in a constant-expression
error: ‘.’ cannot appear in a constant-expression
At global scope:
error: template argument 2 is invalid

是GCC中预期的行为还是错误？

Is that the expected behavior or a bug in GCC?

编辑

正如有人指出，我使用lambda表达式不正确，

As someone pointed out, I'm using lambda expressions incorrectly as they return an instance of the anonymous struct they are referring to.

但是，修复该错误并不能解决问题。我得到 lambda表达式在未评估的上下文错误下面的代码：

However, fixing that error does not solve the problem. I get lambda-expression in unevaluated context error for the following code:

struct A {int x; int y;};
typedef decltype ([](const A lhs, const A &rhs) ->bool {
    return lhs.x < rhs.x;
    }) Comp;
std::set <A, Comp > SetOfA;

推荐答案

std :: set 需要一个类型，而不是一个 表达式，所以它只是你错误地使用。

The 2nd template parameter of std::set expects a type, not an expression, so it is just you are using it wrongly.

您可以这样创建集合：

auto comp = [](const A& lhs, const A& rhs) -> bool { return lhs.x < rhs.x; };
auto SetOfA = std::set <A, decltype(comp)> (comp);

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