＆＃34; Lambda表达式的重要性与C ++ 98＆＃34;不兼容

Question

编译此代码时：

#include <future>
#include <iostream>

int main() {
    std::future<int> result(std::async(
                [](int m, int n) { return m + n; }, 2, 4));

    std::cout << "from main"    << std::endl;
    std::cout << "from asnyc: " << result.get() << std::endl;
    return 0;
}

与

clang++ -std=c++11 -stdlib=libc++ -Weverything promises2.cc -o promises2

我收到了警告：

promises2.cc:6:17: warning: lambda expressions are incompatible with C++98
promises2.cc:5:29: warning: local type '(lambda at promises2.cc:6:17)' as
      template argument is incompatible with C++98

这些是警告，但它们是否应该是可能出错的信号？ Aren＆lt; lambda 认为与C ++ 98不兼容？这些警告应该告诉我的意外是什么？