如何使用带有多个分隔符的istringstream

Hi I'd like to ask how to parse multiple floats, separated by "/" and spaces, from a string.

嗨，我想问一下如何从字符串中解析多个浮点数，用“/”和空格分隔。

The text format from the file is "f 1/1/1 2/2/2 3/3/3 4/4/4" I need to parse every integer from this line of text into several int variables, which are then used to construct a "face" object(see below).

文件中的文本格式是“f 1/1/1 2/2/2 3/3/3 4/4/4”我需要将这行文本中的每个整数解析成几个int变量，然后使用这些变量构造一个“面子”对象（见下文）。

int a(0),b(0),c(0),d(0),e(0);
int t[4]={0,0,0,0};
//parsing code goes here
faces.push_back(new face(b,a,c,d,e,t[0],t[1],t[2],t[3],currentMaterial));

I could do it with sscanf(), but I've been warn away from that by my uni lecturer, so I am looking for an alternative. I am also not allowed other 3rd party libraries, including boost.

我可以用sscanf（）做到这一点，但我已经被我的uni讲师警告过了，所以我正在寻找另一种选择。我也不允许其他第三方库，包括boost。

Regular expressions and parsing with stringstream() have been mentioned, but I don't really know much about either, and would appreciate some advice.

已经提到了正则表达式和使用stringstream（）进行解析，但我对这两者都不太了解，并且会很感激一些建议。

4 个解决方案

#1

If you're reading the file with std::ifstream, there's no need for std::istringstream in the first place (although using the two is very similar because they inherit from the same base class). Here's how to do it with std::ifstream:

如果你正在使用std :: ifstream读取文件，那么首先不需要std :: istringstream（尽管使用两者非常相似，因为它们从相同的基类继承）。以下是使用std :: ifstream的方法：

ifstream ifs("Your file.txt");
vector<int> numbers;

while (ifs)
{
    while (ifs.peek() == ' ' || ifs.peek() == '/')
        ifs.get();

    int number;
    if (ifs >> number)
        numbers.push_back(number);
}

#2

Taking into account your example f 1/1/1 2/2/2 3/3/3 4/4/4 what you need to read is: char int char int char int int char int char int int char int char int

考虑到您的示例f 1/1/1 2/2/2 3/3/3 4/4/4您需要阅读的内容是：char int char int char int int char int char int int char int char int

To do this:

去做这个：

istringstream is(str);

istringstream是（str）;

char f, c;
int d[12];

bool success = (is >> f) && (f == 'f') 
            && (is >> d[0])  && (is >> c) && (c == '/') 
            && (is >> d[1])  && (is >> c) && (c == '/') && 
            .....  && (is >> d[11]);

#3

The way I would do this is to change the interpretation of space to include the other separators. If I were to get fancy I would use different std::ostream objects, each with a std::ctype<char> facet set up to deal with one separator, and use a shared std::streambuf.

我这样做的方法是改变空间的解释以包括其他分隔符。如果我想要使用，我将使用不同的std :: ostream对象，每个对象都设置为std :: ctype facet来处理一个分隔符，并使用共享的std :: streambuf。

If you want to make the use of separators explicit you could instead use a suitable manipulator to skip the separator or, if it absent, indicate failure:

如果您想明确使用分隔符，则可以使用合适的操纵器跳过分隔符，如果不存在则表示失败：

template <char Sep>
std::istream& sep(std::istream& in) {
    if ((in >> std::ws).peek() != std::to_int_type(Sep)) {
        in.setstate(std::ios_base::failbit);
    }
    else {
        in.ignore();
    }
    return in;
}

std::istream& (* const slash)(std::istream&) = Sep<'/'>;

The code isn't tested and type on a mobile device, i.e., probably contains small errors. You'd read data like this:

代码未在移动设备上进行测试和键入，即可能包含小错误。你会读到这样的数据：

if (in >> v1 >> v2 >> slash >> v3 /*...*/) {
  deal_with_input(v1, v2, v3);
}

Note: the above use assumes input as

注意：上面的用法假定输入为

1.0 2.0/3.0

i.e. a space after the first value and a slash after the second value.

即第一个值后的空格和第二个值后的斜杠。

#4

You can use boost::split.

你可以使用boost :: split。

Sample example is:

示例示例如下：

string line("test\ttest2\ttest3");
vector<string> strs;
boost::split(strs,line,boost::is_any_of("\t"));

cout << "* size of the vector: " << strs.size() << endl;    
for (size_t i = 0; i < strs.size(); i++)
    cout << strs[i] << endl;

more information here:

and also related:

还有关：

Splitting the string using boost::algorithm::split

使用boost :: algorithm :: split拆分字符串

#1

ifstream ifs("Your file.txt");
vector<int> numbers;

while (ifs)
{
    while (ifs.peek() == ' ' || ifs.peek() == '/')
        ifs.get();

    int number;
    if (ifs >> number)
        numbers.push_back(number);
}

#2