如何声明一个接受lambda的函数？

问题描述：

我在互联网上阅读了很多教程，解释了如何使用标准库（例如 std :: find ）来使用lambdas，而且都非常有趣，我找不到任何解释我如何使用lambda为我自己的函数。

I read on the internet many tutorials that explained how to use lambdas with the standard library (such as std::find), and they all were very interesting, but I couldn't find any that explained how I can use a lambda for my own functions.

例如：

int main()
{
    int test = 5;
    LambdaTest([&](int a) { test += a; });

    return EXIT_SUCCESS;
}

如何声明 LambdaTest ？它的第一个参数的类型是什么？然后，我如何调用传递给它的匿名函数 - 例如 - 10作为参数？

How should I declare LambdaTest? What's the type of its first argument? And then, how can I call the anonymous function passing to it - for example - "10" as its argument?

答

除了lambdas，你可能还想接受函数指针和函数对象，你可能想使用模板来接受任何带有 operator（）的参数。这是std函数找到做的。它看起来像这样：

Given that you probably also want to accept function pointers and function objects in addition to lambdas, you'll probably want to use templates to accept any argument with an operator(). This is what the std-functions like find do. It would look like this:

template<typename Func>
void LambdaTest(Func f) {
    f(10);
}

请注意，此定义不使用任何c ++ 0x功能，因此它完全向后兼容。这只是使用lambda表达式的函数调用，它是c ++ 0x特定的。

Note that this definition doesn't use any c++0x features, so it's completely backwards-compatible. It's only the call to the function using lambda expressions that's c++0x-specific.

如何声明一个接受lambda的函数？

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