今天有朋友跟我说这么干没问题：

void test() {
    const string &s = some_func();

    add_callback([s] {
        do_something_to(s);
    });
}

我的理解，这个 lambda 会被扩展成这样子：

class lambda {
   const string &s;
   operator () {}
};

一旦这个 lambda 离开了 test()，被 capture 的 s 就不对了。

而我朋友认为是这样:

class lambda {
    const string s;
    operatior () {}
};

有熟读标准的说说这个怎么回事吗？
--
FROM 110.81.42.*

#include <QtCore/qdebug.h>
#include <QtCore/qstring.h>

int main()
{
    const QString &s = QString::fromUtf8("hello!");
    qDebug() << &s;
    ([s] {
        qDebug() << &s;
    })();
    return 0;
}

输出

0x7ffe678780a8
0x7ffe678780b0

说明我朋友说的是对的？

【 在 ble (ble) 的大作中提到: 】
: 写个例子跑一下？
--
FROM 110.81.42.*

规范有明确说明。这句：
“The type of such a data member is the referenced type if the entity is a reference to an object”。

ISO_IEC_14882__2020-12.pdf 这个在110页，"7.5.5.3 Captures"这一节的第10条：

10 An entity is captured by copy if
(10.1) — it is implicitly captured, the capture-default is =, and the captured entity is not *this, or
(10.2) — it is explicitly captured with a capture that is not of the form this, & identifier, or & identifier initializer.
For each entity captured by copy, an unnamed non-static data member is declared in the closure type. The
declaration order of these members is unspecified. The type of such a data member is the referenced type
if the entity is a reference to an object, an lvalue reference to the referenced function type if the entity
is a reference to a function, or the type of the corresponding captured entity otherwise. A member of an
anonymous union shall not be captured by copy.
--
FROM 221.218.165.*

也就是说 [s] 一定会在 lambda 内部生成一个 const T 的字段是吗？其中 T 是原始类型？

外部是`T &&`, `const T&`, `volatile T`, `register T`, `union T`

这里一定会执行复制构造函数？

【 在 z16166 (Netguy) 的大作中提到: 】
: 规范有明确说明。这句：
: “The type of such a data member is the referenced type if the entity is a reference to an object”。
: ISO_IEC_14882__2020-12.pdf 这个在110页，"7.5.5.3 Captures"这一节的第10条：
: ...................
--
FROM 112.47.122.*

话说这句const string &s = some_func();

引用一个函数返回的对象安全吗？

【 在 hgoldfish (老鱼) 的大作中提到: 】
: 今天有朋友跟我说这么干没问题：
: void test() {
:     const string &s = some_func();
: ...................
--
FROM 221.182.193.*