重载运算符>>以从字符串创建数组

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英文:

Overloading operator>> to create array from string

问题

我想从一个std::string创建一个std::array

为此,我想重载operator>>

我有以下的测试用例:

std::istream& operator>>(std::istream& is, const std::array<double, 3>& a)
{
    char p1, p2;

    is >> p1;
    // 如果失败,提醒用户

    for (unsigned int i = 1; i < a.size(); ++i) 
    {
        // 忽略/检查是否为数字的一些操作
    }

    is >> p2;
    // 如果失败,提醒用户

    return is;
}

int main()
{
    std::string a = "[1 2 3]";
    std::array<double, 3> arr;
    std::istringstream iss(a);
    iss >> arr;

    return 0;
}

我希望操作符能够检查字符[]是否处于正确的位置,并使用括号内的元素构造数组。

如何检查提取是否成功?如何检查括号内的字符串是否为数字,如果是,如何从中构造数组?

谢谢!

英文:

I would like to create an std::array from a std::string.

For this, I would like to overload the operator&gt;&gt;.

I have the following test case:

std::istream&amp; operator&gt;&gt;(std::istream&amp; is, const std::array&lt;double, 3&gt;&amp; a)
{
    char p1, p2;

    is &gt;&gt; p1;
    // if fail warn the user

    for (unsigned int i = 1; i &lt; a.size(); ++i) 
    {
        // something to ignore/ check if numeric
    }

    is &gt;&gt; p2;
    // if fail warn the user
    
    return is;
}

int main()
{
    std::string a = &quot;[1 2 3]&quot;;
    std::array&lt;double, 3&gt; arr;
    std::istringstream iss (a);
    iss &gt;&gt; arr;
    
    return 0;
}

I would like for the operator to check if the characters [and ] are in the correct place and to construct the array with the elements inside.

How can I do checks if the extraction was successfull? How can I check the string between the parenthesis is numeric and if so construct my array from it?

Kind regards

答案1

得分: 2

我会为你翻译这段代码,以下是翻译的结果:

我会添加一个辅助类,用于检查流中是否存在特定字符,并在存在时将其移除。如果不存在,则设置 failbit。

示例:

#include <cctype>
#include <istream>

template <char Ch, bool SkipWhitespace = false>
struct eater {
    friend std::istream& operator>>(std::istream& is, eater) {
        if /*constexpr*/ (SkipWhitespace) { // constexpr since C++17
            is >> std::ws;
        }
        if (is.peek() == Ch) // 如果预期的字符存在,则移除它
            is.ignore();
        else                 // 否则设置 failbit
            is.setstate(std::ios::failbit);
        return is;
    }
};

然后可以像这样使用它:

std::istream& operator>>(std::istream& is, std::array<double, 3>& a) {
    // 使用 `eater` 类模板,将 `[` 和 `]` 作为模板参数:
    return is >> eater<'['>{} >> a[0] >> a[1] >> a[2] >> eater<']'>{};
}

int main() {
    std::string a = "[1 2 3]";
    std::array<double, 3> arr;
    std::istringstream iss (a);
    // iss.exceptions(std::ios::failbit); // 如果你想在失败时抛出异常

    if(iss >> arr) {
        std::cout << "success\n";
    } else {
        std::cout << "fail\n";
    }
}

演示 中使用 , 作为输入的分隔符。

英文:

I'd add a helper class that checks for a certain character in the stream and removes it if it's there. If it's not, sets the failbit.

Example:

#include &lt;cctype&gt;
#include &lt;istream&gt;

template &lt;char Ch, bool SkipWhitespace = false&gt;
struct eater {
    friend std::istream&amp; operator&gt;&gt;(std::istream&amp; is, eater) {
        if /*constexpr*/ (SkipWhitespace) { // constexpr since C++17
            is &gt;&gt; std::ws;
        }
        if (is.peek() == Ch) // if the expected char is there, remove it
            is.ignore();
        else                 // else set the failbit
            is.setstate(std::ios::failbit);
        return is;
    }
};

And it could then be used like this:

std::istream&amp; operator&gt;&gt;(std::istream&amp; is, std::array&lt;double, 3&gt;&amp; a) {
    // use the `eater` class template with `[` and `]` as template parameters:
    return is &gt;&gt; eater&lt;&#39;[&#39;&gt;{} &gt;&gt; a[0] &gt;&gt; a[1] &gt;&gt; a[2] &gt;&gt; eater&lt;&#39;]&#39;&gt;{};
}

int main() {
    std::string a = &quot;[1 2 3]&quot;;
    std::array&lt;double, 3&gt; arr;
    std::istringstream iss (a);
    // iss.exceptions(std::ios::failbit); // if you want exceptions on failure

    if(iss &gt;&gt; arr) {
        std::cout &lt;&lt; &quot;success\n&quot;;
    } else {
        std::cout &lt;&lt; &quot;fail\n&quot;;
    }
}

Demo where , is used as separator in the input.

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  • 本文由 发表于 2023年1月8日 21:52:24
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