std::filesystem::hard_link_count
STD：：文件系统：：Hard[医]链接[医]数数
| Defined in header | | |
|---|---|---|
| std::uintmax_t hard_link_count( const std::filesystem::path& p std::uintmax_t hard_link_count( const std::filesystem::path& p, std::error_code& ec | (1) | (since C++17) |
返回按路径标识的文件系统对象的硬链接数。p...
非抛出重载返回。static_cast<uintmax_t>(-1)关于错误。
参数
| p | - | path to examine |
|---|---|---|
| ec | - | out-parameter for error reporting in the non-throwing overload |
返回值
的硬链接数p...
例外
不占用std::error_code&参数抛文件系统[医]误差关于基础OS API错误，使用p作为第一个参数和操作系统错误代码作为错误代码参数。std::bad_alloc如果内存分配失败，则可能引发。过载std::error_code&参数，如果OSAPI调用失败，则将其设置为OSAPI错误代码，并执行ec.clear()如果没有错误发生。这个过载
noexcept规格：
noexcept
例
二次
#include <iostream>
#include <filesystem>
namespace fs = std::filesystem;
int main()
{
// On a POSIX-style filesystem, each directory has at least 2 hard links:
// itself and the special member pathname "."
fs::path p = fs::current_path(
std::cout << "Number of hard links for current path is "
<< fs::hard_link_count(p) << '\n';
// each ".." is a hard link to the parent directory, so the total number
// of hard links for any directory is 2 plus number of direct subdirectories
p = fs::current_path() / ".."; // each dot-dot is a hard link to parent
std::cout << "Number of hard links for .. is "
<< fs::hard_link_count(p) << '\n';
}
二次
可能的产出：
二次
Number of hard links for current path is 2
Number of hard links for .. is 3
二次
另见
| create_hard_link (C++17) | creates a hard link (function) |
|---|
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