遍历与升压::文件系统的目录没有抛出异常

本文介绍了遍历与升压::文件系统的目录没有抛出异常的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我要的目录的路径，我想通过它的所有子目录的遍历，顺便收集文件pathes。

I have a path to the directory, and I want to traverse through all of its sub-directories, collecting files' pathes by the way.

namespace fs = boost::filesystem; std::vector<fs::path> traverse_if_directory(fs::path& f) { std::vector<fs::path> result; if (fs::is_directory(f)) { for (fs::recursive_directory_iterator it(f), eit; it != eit; ++it) { if (!fs::is_directory(it->path())) { result.push_back(it->path()); } } } else { result.push_back(f); } return result; }

不幸的是，在横向的中间，我偶然发现一个目录，我没有权利看，并高于code抛出。但很明显，在这种情况下它不是一个例外，我只是去，这种锁定目录跳过。

Unfortunately, in the middle of the traversing, I stumble upon a directory which I have no rights to look at, and the code above throws. But obviously, in this scenario it's not an exception, I should just go on, skipping this locked directory.

但我怎么做到这一点？

But how do I do this?

推荐答案

呵呵，想通了，还有一个办法：

Ha, figured it out, there is a way:

std::vector<fs::path> traverse_if_directory(fs::path& f) { std::vector<fs::path> result; boost::system::error_code ec; if (fs::is_directory(f)) { for ( fs::recursive_directory_iterator it(f, ec), eit; it != eit; it.increment(ec) ) { if (ec) { it.pop(); continue; } if (!fs::is_directory(it->path())) { result.push_back(it->path()); } } } else { result.push_back(f); } return result; }

有是接受类型的输出参数的非抛过载的boost ::系统::错误_ code ，这样我就可以每次递增之后检查如果有任何错误。

There is a non-throwing overload which accepts an output parameter of type boost::system::error_code, so I can just check after each increment if there were any error.