C++ lambda后跟()与lambda没有()

Question

这两段代码有什么区别？

struct HighResClock
{
    typedef long long                               rep;
    typedef std::nano                               period;
    typedef std::chrono::duration<rep, period>      duration;
    typedef std::chrono::time_point<HighResClock>   time_point;
    static const bool is_steady = true;

    static time_point now();
};


namespace
{
    auto g_Frequency = []() -> long long
    {
        std::cout<<"HERE";
        LARGE_INTEGER frequency;
        QueryPerformanceFrequency(&frequency);
        return frequency.QuadPart;
    }();
}

HighResClock::time_point HighResClock::now()
{
    LARGE_INTEGER count;
    QueryPerformanceCounter(&count);
    return time_point(duration(count.QuadPart * static_cast<rep>(period::den) / g_Frequency));
}

int main()
{
    HighResClock c;
    c.now();
    c.now();
    c.now();
}

和

struct HighResClock
{
    typedef long long                               rep;
    typedef std::nano                               period;
    typedef std::chrono::duration<rep, period>      duration;
    typedef std::chrono::time_point<HighResClock>   time_point;
    static const bool is_steady = true;

    static time_point now();
};


namespace
{
    auto g_Frequency = []() -> long long
    {
        std::cout<<"HERE";
        LARGE_INTEGER frequency;
        QueryPerformanceFrequency(&frequency);
        return frequency.QuadPart;
    };
}

HighResClock::time_point HighResClock::now()
{
    LARGE_INTEGER count;
    QueryPerformanceCounter(&count);
    return time_point(duration(count.QuadPart * static_cast<rep>(period::den) / g_Frequency()));
}

int main()
{
    HighResClock c;
    c.now();
    c.now();
    c.now();
}

如果你没有注意到,差异是下面的括号:

auto g_Frequency = []() -> long long
{
    LARGE_INTEGER frequency;
    QueryPerformanceFrequency(&frequency);
    return frequency.QuadPart;
}(); //this bracket here appears in one and not the other..

我问,因为带括号的那个只打印一次"Here",而另一个(没有括号)打印3次.括号是什么意思,它有什么作用？括号中是否有此语法的名称？

Answer 1

()在lambda定义之后立即写入[]{}();将调用lambda,因此结果是lambda的返回类型的类型.

如果省略()后缀,将返回lambda类型(未指定),这基本上是一个可调用的函子.

auto result = []{ return 42; }(); // auto is integer, result has 42 in it  
auto result1 = []{ return 42; }; // auto is some unspecified lambda type  
auto result2 = result1(); // auto is integer, result2 is storing 42` 
......................^^ - this is the bracket you can also put directly after the definition of the lambda