如何测试stringstream运算符>>已解析了一个坏类型并跳过它 [英] How to test whether stringstream operator>> has parsed a bad type and skip it

问题描述

我有兴趣讨论使用 stringstream 来解析多种类型的行的方法。我将从以下行开始：

I am interested in discussing methods for using stringstream to parse a line with multiple types. I would begin by looking at the following line:

"2.832 1.3067 nana 1.678"

现在让我假设我有一个长线有多个字符串和 doubles 。解决这个问题的明显方法是对字符串进行标记，然后检查转换每个字符串。我有兴趣跳过这个第二步，直接使用 stringstream 只找到数字。

Now lets assume I have a long line that has multiple strings and doubles. The obvious way to solve this is to tokenize the string and then check converting each one. I am interested in skipping this second step and using stringstream directly to only find the numbers.

我认为一个好方法是通过读取字符串并检查 failbit 设置，这将是如果我试图解析一个字符串成双。

I figured a good way to approach this would be to read through the string and check if the failbit has been set, which it will if I try to parse a string into a double.

说我有以下代码：

string a("2.832 1.3067 nana 1.678");

 stringstream parser;
 parser.str(a);

 for (int i = 0; i < 4; ++i)
 {
     double b;
     parser >> b;
     if (parser.fail())
     {
         std::cout << "Failed!" << std::endl;
         parser.clear();
     }
     std::cout << b << std::endl;
 }

它将打印出以下内容：

2.832
1.3067
Failed!
0
Failed!
0

我不奇怪，它无法解析字符串， failbit 并解析下一个数字？

I am not surprised that it fails to parse a string, but what is happening internally such that it fails to clear its failbit and parse the next number?

推荐答案

以下代码适用于跳过坏词并收集有效的 double 值

The following code works well to skip the bad word and collect the valid double values

istringstream iss("2.832 1.3067 nana 1.678");
double num = 0;
while(iss >> num || !iss.eof()) {
    if(iss.fail()) {
        iss.clear();
        string dummy;
        iss >> dummy;
        continue;
    }
    cout << num << endl;
}

这里是完全工作的样品。

您的样品差不多正确，检测到格式错误之后流中的无效输入字段

Your sample almost got it right, it was just missing to consume the invalid input field from the stream after detecting it's wrong format

 if (parser.fail()) {
     std::cout << "Failed!" << std::endl;
     parser.clear();
     string dummy;
     parser >> dummy;
 }

在您的情况下，解压缩将尝试从nana用于最后一次迭代，因此输出中的最后两行。

In your case the extraction will try to read again from "nana" for the last iteration, hence the last two lines in the output.

还要注意 iostream :: fail（）以及如何在我的第一个示例中实际测试 iostream :: eof（）。 有一个众所周知的Q& A ，为什么简单测试EOF作为循环条件被认为是错误的。它回答很好，当遇到意外/无效值时如何打破输入循环。但是如何跳过/忽略无效的输入字段在那里没有解释（并没有要求）。

Also note the trickery about iostream::fail() and how to actually test for iostream::eof() in my 1st sample. There's a well known Q&A, why simple testing for EOF as a loop condition is considered wrong. And it answers well, how to break the input loop when unexpected/invalid values were encountered. But just how to skip/ignore invalid input fields isn't explained there (and wasn't asked for).

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