Compares the lexical representations of the path and another path.
root_name().native().compare(p.root_name().native())
is nonzero, returns that value. 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.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.compare(path(str))
.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.
#include <iostream> #include <filesystem> namespace fs = std::filesystem; void demo(int rc, fs::path p1, fs::path p2) { if(rc < 0) std::cout << p1 << " < " << p2 << '\n'; else if(rc > 0) std::cout << p1 << " > " << p2 << '\n'; else if(rc == 0) std::cout << p1 << "==" << p2 << '\n'; } int main() { fs::path p1 = "/a/b/"; // as if "a/b/." for lexicographical iteration fs::path p2 = "/a/b/#"; demo(p1.compare(p2), p1, p2); demo(p1.compare("a/b/_"), p1, "a/b/_"); }
Output:
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 |
lexicographically compares two paths (function) |
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