c++ - 为什么在 lambda 中获取指向成员函数的指针会在 VS2013 上触发警告

Question

在 visual studio 2013(安装了更新 5)上编译代码并使用默认警告级别 (/W3)

#include <iostream>

class Foo {
public:
    void Thunk()
    {
        auto lambda = [](Foo* f, const char* msg) {
            auto pfn = &Foo::Print;  // complained this statement
            (*f.*pfn)(msg);
        };

        lambda(this, "abc");
    }

    void Print(const char* msg)
    {
        std::cout << msg << std::endl;
    }
};

int main()
{
    Foo foo;

    foo.Thunk();

    return 0;
}

会触发编译警告:

Line 16: warning C4573: the usage of 'Foo::Print' requires the compiler to capture 'this' but the current default capture mode does not allow it

但是，如果我在 Visual Studio 2017 (ver15.4.2) 上编译这段代码，这个警告就会消失；并使用/W4 警告级别编译仍然无法重现警告。

这是 Visual Studio 2013 的实现错误吗？

Answer 1

发生这种情况是因为您在 lambda 声明中有默认的捕获选项。只需检查 Lambda 捕获 部分 here

我认为你应该用 [=]

来改变它

auto lambda = [=](Foo* f, const char* msg) {
    auto pfn = &Foo::Print;  // complained this statement
        (*f.*pfn)(msg);
};

从上面的链接中，这是可能的捕获选项:

captures - a comma-separated list of zero or more captures, optionally beginning with a capture-default.

• [a,&b] where a is captured by copy and b is captured by reference.
• [this] captures the current object (*this) by reference
• [&] captures all automatic variables used in the body of the lambda by reference and current object by reference if exists
• [=] captures all automatic variables used in the body of the lambda by copy and current object by reference if exists
• [] captures nothing

A variable can be used without being captured if it does not have automatic storage duration (i.e. it is not a local variable or it is static or thread local) or if it is not odr-used in the body of the lambda.