为什么我不能在C ++ 11中创建一个lambda的向量？ [英] Why can't I create a vector of lambda in C++11?

问题描述

我试图创建一个lambda向量，但失败：

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，它编译精细。但第＃3行会导致编译错误：

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()>)'

我不想要一个函数指针的向量或函数对象的向量。然而，封装了 lambda表达式的函数对象的向量将对我有用。这是可能的吗？

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?

推荐答案

每个lambda都有不同的类型 - 即使他们有相同的签名。如果你想这样做，你必须使用运行时封装容器，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; });

我不想要一个函数指针向量或函数向量
对象。

I don't want a vector of function pointers or vector of function objects.

好吧，我想要一个很好的高薪工作，我的梦想，但这不会发生。哦，我也想要一个非常有魅力的妻子，他可以做饭和清洁，因为我真的不能。

Well, I would like a nice high-paying job with a hundred-man developer team to develop all my dreams, but that's not gonna happen either. Oh, and I'd also like a very attractive wife who can cook and clean, since I really can't.

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