c++ - 存储在类中的闭包中通过引用捕获的临时对象的生命周期

Question

考虑以下代码片段:

struct foo { };

template <typename F>
struct impl : F
{
    impl(F&& f) : F{std::move(f)} { }
    auto get() { return (*this)(); }
};

template <typename X>
auto returner(X&& x)
{
    return impl{[&x]{ return x; }};
//               ^~
}

int main()
{
    auto x = returner(foo{}).get();
}

live example on wandbox.org

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• 是否保证 foo{} 在 returner(foo{}).get() 表达式的整个持续时间内都有效？

• 或者 foo{} 是否只对 returner(foo{}) 有效，从而导致未定义的行为调用 impl::get()?

---

标准在 [class.temporary] 中说:

Temporary objects are destroyed as the last step in evaluating the full-expression that (lexically) contains the point where they were created.

在[intro.execution]

A full-expression is

• an unevaluated operand,

• a constant-expression,

• an init-declarator or a mem-initializer, including the constituent expressions of the initializer,

• an invocation of a destructor generated at the end of the lifetime of an object other than a temporary object ([class.temporary]), or

• an expression that is not a subexpression of another expression and that is not otherwise part of a full-expression.

我不确定与 foo{} 相关的 full-expression 是 returner(foo{}) 还是 returner(foo{}).get().

Answer 1

这里的重要部分是:

A full-expression is [...] an expression that is not a subexpression of another expression and that is not otherwise part of a full-expression.

所以在 returner(foo{}).get() 中，returner(foo{}) 是表达式 returner(foo{} ).get()，所以它不是一个完整的表达式。因此:

Is it guaranteed that foo{} will be alive for the entire duration of the returner(foo{}).get() expression?

是的。