为什么我不能在C ++ 11中创建Lambda向量(相同类型)?

问题描述：

我试图创建lambda向量，但失败了:

I was trying to create a vector of lambda, but failed:

auto ignore = [&]() { return 10; };  //1
std::vector<decltype(ignore)> v;     //2
v.push_back([&]() { return 100; });  //3

直到第2行，它可以正常编译.但是第3行给出了编译错误:

Up to line #2, it compiles fine. But the line#3 gives compilation error:

错误:没有匹配函数调用'std :: vector< main()::< lambda()>> :: push_back(main()::< lambda()>)'

error: no matching function for call to 'std::vector<main()::<lambda()>>::push_back(main()::<lambda()>)'

我不需要功能指针的向量或功能对象的向量.但是，封装 real lambda表达式的功能对象的矢量对我来说将是有效的.这可能吗?

I don't want a vector of function pointers or vector of function objects. However, vector of function objects which encapsulate real lambda expressions, would work for me. Is this possible?

答

每个lambda都有不同类型—即使它们具有相同的签名.如果要执行类似的操作，则必须使用运行时封装容器，例如std::function.

Every lambda has a different type—even if they have the same signature. You must use a run-time encapsulating container such as std::function if you want to do something like that.

例如:

std::vector<std::function<int()>> functors;
functors.push_back([&] { return 100; });
functors.push_back([&] { return  10; });

为什么我不能在C ++ 11中创建Lambda向量(相同类型)?

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