std::filesystem::path::compare (3) - Linux Man Pages
int compare( const path& p ) const noexcept; (1) (since C++17)
int compare( const string_type& str ) const; (2) (since C++17)
int compare( std::basic_string_view<value_type> str ) const;
int compare( const value_type* s ) const; (3) (since C++17)
Compares the lexical representations of the path and another path.
1) If root_name().native().compare(p.root_name().native()) is nonzero, returns that value.
Otherwise, if has_root_directory() != p.has_root_directory(), returns a value less than zero if has_root_directory() is false and a value greater than zero otherwise.
Otherwise returns a value less than, equal to or greater than 0 if the relative portion of the path (relative_path()) is respectively lexicographically less than, equal to or greater than the relative portion of p (p.relative_path()). Comparison is performed element-wise, as if by iterating both paths from begin() to end() and comparing the result of native() for each element.
2) Equivalent to compare(path(str)).
3) Equivalent to compare(path(s)).
p - a path to compare to
str - a string or string view representing path to compare to
s - a null-terminated string representing path to compare to
A value less than 0 if the path is lexicographically less than the given path.
A value equal to 0 if the path is lexicographically equal to the given path.
A value greater than 0 if the path is lexicographically greater than the given path.
For two-way comparisons, binary_operators may be more suitable.
// Run this code
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR Applied to Behavior as published Correct behavior
LWG_2936 C++17 compared all path elements directly root name and root directory handled separately
operator!= lexicographically compares two paths