粘性自定义流操纵器 [英] Sticky custom stream manipulator

问题描述

如何实现我自己的自定义流操纵器，使其是粘性的。例如，我想将整数转换为二进制，以便：

How do I implement my own custom stream manipulator so that it is sticky. For example, I want to convert integers to binary such that:

cout << "decimal of 4: " <<  4 
     << "\ndecimal of 4: " << 4 
     << binary << "\nbinary of 4: " << 4 
     << "\nbinary of 4: " << 4 
     << nobinary << "\ndecimal of 4: " << 4 
     << "\ndecimal of 4: " << 4 << endl;

会返回：

decimal of 4: 4
decimal of 4: 4
binary of 4: 100
binary of 4: 100
decimal of 4: 4
decimal of 4: 4

推荐答案

有点涉及。为了使其易于理解，我将从基本的东西开始：为用户定义的类型使用自定义格式化标志。

Doing the whole things is a bit involved. To make it comprehensible, I'll start with the basic stuff: Using custom formatting flags for user-defined types. Custom formatting of integers will follow below.

IOStream类从 std :: ios_base 中获得[间接]两个数据存储： std :: ios_base :: iword（）和 std :: ios_base :: pword（） int s和 void * 。使用 std :: ios_base :: pword（）保存分配的内存是不平凡的，幸运的是，这个相对简单的用例不需要。要使用这两个函数都返回一个非< - c $ c> const 引用对应的类型，你通常使用 std :: ios_base :: xalloc ）一次，并在您需要访问自定义格式标记时使用它。当您使用 iword（）或 pword（）访问一个值时，它将初始化为零。为了把事情放在一起，这里是一个小程序演示这个：

The IOStream classes derive [indirectly] from std::ios_base which provides two stores for data: std::ios_base::iword() and std::ios_base::pword() for ints and void*, respectively. Maintaining allocated memory stored with std::ios_base::pword() is non-trivial and, fortunately, not needed for this relatively simple use-case. To use these function which both return a non-const reference to the corresponding type, you normally allocate an index using std::ios_base::xalloc() once in your program and use it whenever you need to access your custom formatting flags. When you access a value with iword() or pword() initially it will be zero initialized. To put things together, here is a small program demonstrating this:

#include <iostream>

static int const index = std::ios_base::xalloc();

std::ostream& custom(std::ostream& stream) {
    stream.iword(index) = 1;
    return stream;
}

std::ostream& nocustom(std::ostream& stream) {
    stream.iword(index) = 0;
    return stream;
}

struct mytype {};
std::ostream& operator<< (std::ostream& out, mytype const&) {
    return out << "custom-flag=" << out.iword(index);
}

int main()
{
    std::cout << mytype() << '\n';
    std::cout << custom;
    std::cout << mytype()  << '\n';
    std::cout << nocustom;
    std::cout << mytype() << '\n';
}

现在， int like 4 不是用户定义类型，并且已经为这些类型定义了输出运算符。幸运的是，你可以使用facet，更具体地使用 std :: num_put< char> 来自定义整数的格式。现在，为此，您需要执行一些步骤：

Now, an int like 4 isn't a user-define type and there is already an output operator defined for these. Fortunately, you can customize the way integers get formatted using facets, more specifically using std::num_put<char>. Now, to do so you need to do a number of steps:

1. 从 std :: num_put< ; char> 并覆盖您要赋予其专门行为的 do_put（）成员。


2. 创建 ）具有新 std :: locale 的流。

1. Derive a class from std::num_put<char> and override the do_put() members you want to give specialized behavior to.
2. Create a std::locale object using the newly create facet.
3. std::ios_base::imbue() the stream with the new std::locale.

$ b b

为了让用户更好用，你可能需要用一个合适的 std :: num_put< end来创建一个新的 std :: locale ; char> facet当使用操纵器时。但是，在这样做之前，让我们先创建一个合适的facet：

To make things nicer for the user, you might want to conjure up a new std::locale with a suitable std::num_put<char> facet when the manipulator is used. However, before doing so, let's start off with creating a suitable facet:

#include <bitset>
#include <iostream>
#include <limits>
#include <locale>

static int const index = std::ios_base::xalloc();

class num_put
    : public std::num_put<char>
{
protected:
    iter_type do_put(iter_type to,
                     std::ios_base& fmt,
                     char_type fill,
                     long v) const
    {
        if (!fmt.iword(index)) {
            return std::num_put<char>::do_put(to, fmt, fill, v);
        }
        else {
            std::bitset<std::numeric_limits<long>::digits> bits(v);
            size_t i(bits.size());
            while (1u < i && !bits[i - 1]) {
                --i;
            }
            for (; 0u < i; --i, ++to) {
                *to = bits[i - 1]? '1': '0';
            }
            return to;
        }
    }
#if 0
    // These might need to be added, too:
    iter_type do_put(iter_type, std::ios_base&, char_type,
                     long long) const;
    iter_type do_put(iter_type, std::ios_base&, char_type,
                     unsigned long) const;
    iter_type do_put(iter_type, std::ios_base&, char_type,
                     unsigned long long) const;
#endif
};

std::ostream& custom(std::ostream& stream) {
    stream.iword(index) = 1;
    return stream;
}

std::ostream& nocustom(std::ostream& stream) {
    stream.iword(index) = 0;
    return stream;
}

int main()
{
    std::locale loc(std::locale(), new num_put);
    std::cout.imbue(loc);
    std::cout << 13 << '\n';
    std::cout << custom;
    std::cout << 13  << '\n';
    std::cout << nocustom;
    std::cout << 13 << '\n';
}

有点丑的是必须 imbue（）使用自定义操纵器的自定义 std :: locale 为了摆脱这个，我们可以确保自定义facet安装在使用的 std :: locale ，如果不是，只是安装它，当设置标志：

What is a bit ugly is that it necessary to imbue() the custom std::locale to use the custom manipulator. To get rid of this, we can just make sure the custom facet is installed in the used std::locale and, if it is not, just install it when setting the flag:

std::ostream& custom(std::ostream& stream) {
    if (!stream.iword(index)
        && 0 == dynamic_cast<num_put const*>(
                &std::use_facet<std::num_put<char> >(stream.getloc()))) {
        stream.imbue(std::locale(stream.getloc(), new num_put));
    }
    stream.iword(index) = 1;
    return stream;
}

现在剩下的还是重写不同的 do_put（）成员与各种无符号类型以及 long long 这是一个练习。

What is now left is to also override the different do_put() members to work properly with the various unsigned types and with long long but this is left as an exercise.

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