c++ - 在同一语句中返回传入的临时值并从中读取是否安全？

Question

我只是写了这篇文章，没有想太多。它似乎工作正常，但我不确定它是否绝对安全。

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 Buffer 将一直存在，直到整个 cout 语句完成。是这样吗，还是会在此之前超出范围，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 lexicallypart of the full-expression. For example, subexpressions involved in evaluating default arguments (8.3.6) areconsidered to be created in the expression that calls the function, not the expression that defines the defaultargument. — end note ]

不过，这并不意味着它特别好 - 将 Foo().print() 的结果绑定(bind)到 太容易了>char const* 变量，它将在下一个完整表达式中变成悬空指针。