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# std::experimental::filesystem::path::compare

From cppreference.com
< cpp‎ | experimental‎ | fs‎ | path

int compare( const path& p ) const; | (1) | (filesystem TS)
int compare( const string_type& str ) const; | (2) | (filesystem TS)
int compare( const value_type* s ) const; | (3) | (filesystem TS)

Compares the lexical representations of the path and another path.

1) Returns a value less than, equal to or greater than 0 if the native representation of the path (native()) is respectively lexicographically less than, equal to or greater than the native representation of `p` (p.native()). Comparison is performed element-wise, as if by iterating both paths from begin() to end().
2) Equivalent to compare(path(str)).
3) Equivalent to compare(path(s)).

### [edit] Parameters

p | - | a path to compare to
str | - | a string representing path to compare to
s | - | a null-terminated string representing path to compare to

### [edit] Return value

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.

### [edit] Exceptions

1)
`noexcept` specification:
noexcept
2-3) (none)

### [edit] Notes

For two-way comparisons, binary operators may be more suitable.

### [edit] Example

Run this code

    #include <iostream>
    #include <experimental/filesystem>
    namespace fs = std::experimental::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:

    "/a/b/" > "/a/b/#"
    "/a/b/" < "a/b/_"

### [edit] See also

operator==operator!=operator<operator<=operator>operator>= | lexicographically compares two paths (function) [edit]

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