c++ - 从空缓冲区构造 `std::ostream` 是否有效？

Question

考虑以下几点:

std::ostream out(nullptr);

这合法且定义明确吗？

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如果我现在这样做呢:

out << "hello world\n";

这是否合法且定义明确？如果是这样，大概是一种无操作？

Answer 1

是的，实例化该流是合法且定义明确的。您可以安全地将其与另一个流交换，或者稍后给它一个新指针(这次指向现有缓冲区)。输出操作本身确实是空操作。

原因如下:

1. 构造没有非空前置条件，只有这个后置条件:

   [C++11: 27.7.3.2/2]: Postcondition: rdbuf() == sb.

2. 有趣的是，它明确指出不得对 sb 执行任何操作。在构造函数中:

   [C++11: 27.7.3.2/4]: Remarks: Does not perform any operations on rdbuf().

3. 但也要注意:

   [C++11: 27.7.3.2/1]: Effects: Constructs an object of class basic_ostream, assigning initial values to the base class by calling basic_ios<charT,traits>::init(sb) (27.5.5.2).

4. 那个init(sb) call具有设置badbit的效果直播时sb为空:

   [C++11: 27.5.5.2/3]: Postconditions: The postconditions of this function are indicated in Table 128.

   [C++11: Table 128]: [..] rdstate(): goodbit if sb is not a null pointer, otherwise badbit. [..]

5. 输出操作会导致相当于取消引用空指针的操作:

   [C++11: 27.7.3.1/2]: Two groups of member function signatures share common properties: the formatted output functions (or inserters) and the unformatted output functions. Both groups of output functions generate (or insert) output characters by actions equivalent to calling rdbuf()->sputc(int_type). They may use other public members of basic_ostream except that they shall not invoke any virtual members of rdbuf() except overflow(), xsputn(), and sync().

   除了它从来没有走到这一步，因为对于basic_ostream::sentry build :

   [C++11: 27.7.3.4/3]: If, after any preparation is completed, os.good() is true, ok_ == true otherwise, ok_ == false.

   对于explicit operator basic_ostream::sentry::bool() const; :

   [C++11: 27.7.3.4/5]: Effects: Returns ok_.

   和:

   [C++11: 27.7.3.7/1]: Each unformatted output function begins execution by constructing an object of class sentry. If this object returns true, while converting to a value of type bool, the function endeavors to generate the requested output. [..]

   ...这意味着当 badbit 时根本不发生任何输出操作。已经设置好了。

This was also the case in C++03.