修剪空白/转换迭代器
我想编写两个单行函数，它们将修剪字符串左侧和右侧的空格。左侧没有问题：
void trimLeft(string &s) {
s.erase(s.begin(), find_if(s.begin(), s.end(), (int (*)(int))isgraph));
}
但是当我尝试右侧类似的操作时：
void trimRight(string &s) {
s.erase(find_if(s.rbegin(), s.rend(), (int (*)(int))isgraph), s.end());
}
我遇到了一些编译器错误。问题是我必须将reverse_iterator（由find_if返回）转换为普通迭代器。如何做到这一点？
I wanted to write two one-line functions which will trim white spaces from left and right side of a string. Left side weren't problem:
void trimLeft(string &s) {
s.erase(s.begin(), find_if(s.begin(), s.end(), (int (*)(int))isgraph));
}
But when I tried something similar for right side:
void trimRight(string &s) {
s.erase(find_if(s.rbegin(), s.rend(), (int (*)(int))isgraph), s.end());
}
I had some compiler errors. The problem is that I must convert reverse_iterator (which is returned by find_if) to normal iterator. How to do this?
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您可以使用
base()成员函数从其reverse_iterator恢复底层迭代器。You can use the
base()member function to recover the underlying iterator from its reverse_iterator.查看文档，似乎 string::erase 应该采用reverse_iterators 就可以了。问题是您混合了反向和正向迭代器。所以尝试
编辑：不。
Looking at the documentation, it seems like string::erase should take reverse_iterators just fine. The problem is that you're mixing reverse and forward iterators. So try
Edit: Nope.
使用 boost::trim 吗？
http://www.boost.org/doc /libs/1_47_0/doc/html/string_algo/usage.html#id2895820
Use boost::trim?
http://www.boost.org/doc/libs/1_47_0/doc/html/string_algo/usage.html#id2895820