return 传入的临时值并在同一语句中从中读取是否安全？

Question

我只是没有想太多就写了这篇文章。貌似还行，但不确定是否绝对安全。

class Foo
{
    struct Buffer
    {
        char data [sizeof ("output will look like this XXXX YYYY ZZZZ")];
    };

    const char * print (const char * format = DEFUALT_FORMAT, Buffer && buf = Buffer ())
    {
        sort_of_sprintf_thing (format, buf .data, sizeof (buf.data), ...);
        return buf .data;
    }
};

std :: cout << Foo () .print ();

所以我认为语义是临时缓冲区将一直存在，直到整个cout语句完成。是这样吗，还是在那之前它会超出范围，在这种情况下这是 UB？

Answer 1

是的，您的代码定义明确。

[class.temporary]

3 - [...] Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. [...]

[intro.execution]

11 - [ Note: The evaluation of a full-expression can include the evaluation of subexpressions that are not lexically part of the full-expression. For example, subexpressions involved in evaluating default arguments (8.3.6) are considered to be created in the expression that calls the function, not the expression that deﬁnes the default argument. — end note ]

这并不意味着它特别好，但是 - 将 Foo().print() 的结果绑定到 char const* 变量太容易了, 这将在下一个完整表达式中成为悬空指针。

Answer 2

代码有问题，问题不在调用站点上，而是在 print 函数上。您正在使用 rvalue（唯一会绑定到 rvalue-reference 的东西）并返回指向其内部结构的指针，这是 Undefined 的秘诀行为（如果用户取消引用返回的 const char*）。

在您的特定示例中，Foo() 临时变量将存在足够长的时间，但此代码很容易让客户端将 const char* 存储在完整表达式之外并导致未定义的行为。

return 传入的临时值并在同一语句中从中读取是否安全？

Is it safe to return a passed-in temporary value and read from it in the same statement?

c++

scope

undefined-behavior

[class.temporary]

[intro.execution]