C++ 在两个迭代器之间获取`std::string`的子字符串
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Getting a substring of a `std::string` between two iterators
提问by ely
I have a program that's expected to take as input a few options to specify probabilities in the form (p1, p2, p3, ... ). So the command line usage would literally be:
我有一个程序，它希望将一些选项作为输入，以指定形式 (p1, p2, p3, ... ) 中的概率。所以命令行用法实际上是：
./myprog -dist (0.20, 0.40, 0.40)
I want to parse lists like this in C++ and I am currently trying to do it with iterators for the std::string type and the split function provided by Boost.
我想在 C++ 中解析这样的列表，我目前正在尝试使用 std::string 类型的迭代器和 Boost 提供的拆分函数来完成它。
// Assume stuff up here is OK
std::vector<string> dist_strs; // Will hold the stuff that is split by boost.
std::string tmp1(argv[k+1]); // Assign the parentheses enclosed list into this std::string.
// Do some error checking here to make sure tmp1 is valid.
boost::split(dist_strs, <what to put here?> , boost::is_any_of(", "));
Note above the <what to put here?>part. Since I need to ignore the beginning and ending parentheses, I want to do something like
注意上面的<what to put here?>部分。由于我需要忽略开头和结尾的括号，我想做类似的事情
tmp1.substr( ++tmp1.begin(), --tmp1.end() )
but it doesn't look like substrworks this way, and I cannot find a function in the documentation that works to do this.
但它看起来不像这样substr工作，而且我在文档中找不到可以执行此操作的函数。
One idea I had was to do iterator arithmetic, if this is permitted, and use substrto call
我的一个想法是进行迭代器算术，如果允许的话，并用于substr调用
tmp1.substr( ++tmp1.begin(), (--tmp1.end()) - (++tmp1.begin()) )
but I wasn't sure if this is allowed, or if it is a reasonable way to do it. If this isn't a valid approach, what is a better one? ...Many thanks in advance.
但我不确定这是否被允许，或者这是否是一种合理的方式。如果这不是一种有效的方法，那么什么是更好的方法？...提前谢谢了。
回答by Pubby
std::string's constructor should provide the functionality you need.
std::string的构造函数应该提供您需要的功能。
std::string(tmp1.begin() + 1, tmp1.end() - 1)
