限制 boost::options_description 中默认值的 std::cout 精度

发布于 2024-08-10 19:59:18 字数 622 浏览 14 评论 0原文

当我构造一个 boost::options_description 实例时，例如，

options.add_options()
  ("double_val", value(&config.my_double)->default_value(0.2), "it's a double");

稍后想要自动输出可用于我的程序的选项，并

std::cout << options << std::endl;

以太高的精度显示默认值 0.2 ，这实际上会在我有时使我的输出变得混乱长变量名：

--double_val (=0.20000000000000001) it's a double

不幸的是，之前对 std::cout. precision 的调用没有帮助：

cout.precision(5);
std::cout << options << std::endl;

这仍然导致相同的输出：/

您对如何将默认值的显示限制在更少的位置有什么想法吗？

此致，基督教

原文

When I construct a boost::options_description instance like

options.add_options()
  ("double_val", value(&config.my_double)->default_value(0.2), "it's a double");

and later want to have the automated output of the options that are available for my program, and put

std::cout << options << std::endl;

the default value 0.2 is shown with way too high precision, which effectively clutters my output when I have long variable names:

--double_val (=0.20000000000000001) it's a double

unfortunately, a prior call to std::cout.precision did not help:

cout.precision(5);
std::cout << options << std::endl;

this leads still to the same output :/

Do you have any ideas on how to limit the display of the default value to less positions?

Best regards,
Christian

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甜味超标? 2024-08-17 19:59:18

来自boost/program_options/value_semantic.hpp：

    /** Specifies default value, which will be used
        if none is explicitly specified. The type 'T' should
        provide operator<< for ostream.
    */
    typed_value* default_value(const T& v)
    {
        m_default_value = boost::any(v);
        m_default_value_as_text = boost::lexical_cast<std::string>(v);
        return this;
    }

    /** Specifies default value, which will be used
        if none is explicitly specified. Unlike the above overload,
        the type 'T' need not provide operator<< for ostream,
        but textual representation of default value must be provided
        by the user.
    */
    typed_value* default_value(const T& v, const std::string& textual)
    {
        m_default_value = boost::any(v);
        m_default_value_as_text = textual;
        return this;
    }

所以实现非常简单（Boost 从来都不是确定的事情！）。尝试重新配置 ostream 以使格式按照您想要的方式显示是行不通的，因为默认值只是转换为独立 ostringstream 中的字符串（在 lexical_cast 内））。

因此，一个简单的解决方法是将所需的字符串表示形式添加为 default_value 的第二个参数。然后你可以让它按照你想要的方式打印（包括根本不打印，如果你传递一个空字符串）。像这样：

value(&config.my_double)->default_value(0.2, "0.2")

完成同样事情的更“企业”方法是实现您自己的类型，该类型将包装 double，用于 config.my_double，并提供构造from 并强制转换为 double，以及您自己的 ostream&运算符<< 与您想要的格式完全相同。但是，我不建议使用这种方法，除非您正在编写一个需要通用性的库。

来自 Boost Lexical Cast 的注释：

之前版本的 lexical_cast
使用默认流精度
读取和写入浮点数
数字。对于具有
相应的专业
std::numeric_limits，当前
版本现在选择精度
匹配。

From boost/program_options/value_semantic.hpp:

    /** Specifies default value, which will be used
        if none is explicitly specified. The type 'T' should
        provide operator<< for ostream.
    */
    typed_value* default_value(const T& v)
    {
        m_default_value = boost::any(v);
        m_default_value_as_text = boost::lexical_cast<std::string>(v);
        return this;
    }

    /** Specifies default value, which will be used
        if none is explicitly specified. Unlike the above overload,
        the type 'T' need not provide operator<< for ostream,
        but textual representation of default value must be provided
        by the user.
    */
    typed_value* default_value(const T& v, const std::string& textual)
    {
        m_default_value = boost::any(v);
        m_default_value_as_text = textual;
        return this;
    }

So the implementation is dead simple (never a sure thing with Boost!). Trying to reconfigure your ostream to make the formatting come out as you want won't work, because the default value just gets converted to a string in a standalone ostringstream (inside lexical_cast).

So a simple workaround is to add your desired string representation as a second argument to default_value. Then you can make it print however you want (including not at all, if you pass an empty string). Like so:

value(&config.my_double)->default_value(0.2, "0.2")

The more "enterprisey" way to accomplish the same thing would be to implement your own type which would wrap double, be used for config.my_double, and provide construction from and coercion to double, and your very own ostream& operator<< with exactly the formatting you desire. I don't suggest this approach, however, unless you're writing a library that demands generality.

From the Boost Lexical Cast notes:

The previous version of lexical_cast
used the default stream precision for
reading and writing floating-point
numbers. For numerics that have a
corresponding specialization of
std::numeric_limits, the current
version now chooses a precision to
match.

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独自唱情﹋歌 2024-08-17 19:59:18

为了避免手动引用：

#define QUOTE(x) #x
#define stringize(x) QUOTE(x)

#define MY_DOUBLE_DEFAULT 0.2

value(&config.my_double)->default_value(MY_DOUBLE_DEFAULT, stringize(MY_DOUBLE_DEFAULT))

To avoid having to quote by hand:

#define QUOTE(x) #x
#define stringize(x) QUOTE(x)

#define MY_DOUBLE_DEFAULT 0.2

value(&config.my_double)->default_value(MY_DOUBLE_DEFAULT, stringize(MY_DOUBLE_DEFAULT))

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