std::collate::transform, do_transform
From cppreference.com
Defined in header
<locale>
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public:
string_type transform( const CharT* low, const CharT* high ) const; |
(1) | |
protected:
virtual string_type do_transform( const CharT* low, const CharT* high ) const; |
(2) | |
1) Public member function, calls the protected virtual member function
do_transform
of the most derived class.
2) Converts the character sequence
[low, high)
to a string that, compared lexicographically (e.g. with operator<
for strings) with the result of calling transform()
on another string, produces the same result as calling do_compare() on the same two strings.
Contents |
[edit] Parameters
low | - | pointer to the first character in the sequence to transform |
high | - | one past the end pointer for the sequence to transform |
[edit] Return value
The string transformed so that lexicographic comparison of the transformed strings may be used instead of collating of the originals. In the "C" locale, the returned string is the exact copy of [low, high)
. In other locales, the contents of the returned string are implementation-defined, and the size may be considerably longer.
[edit] Notes
In addition to the the use in collation, the implementation-specific format of the transformed string is known to std::regex_traits<>::transform_primary, which is able to extract the equivalence class information.
[edit] Example
Run this code
#include <iostream> #include <iomanip> #include <locale> int main() { std::locale::global(std::locale("sv_SE.utf8")); auto& f = std::use_facet<std::collate<wchar_t>>(std::locale()); std::wstring in1 = L"\u00e4ngel"; std::wstring in2 = L"\u00e5r"; std::wstring out1 = f.transform(&in1[0], &in1[0] + in1.size()); std::wstring out2 = f.transform(&in2[0], &in2[0] + in2.size()); std::wcout << "In the Swedish locale: "; if(out1 < out2) std::wcout << in1 << " before " << in2 << '\n'; else std::wcout << in2 << " before " << in1 << '\n'; std::wcout << "In lexicographic comparison: "; if(in1 < in2) std::wcout << in1 << " before " << in2 << '\n'; else std::wcout << in2 << " before " << in1 << '\n'; }
Output:
In the Swedish locale: år before ängel In lexicographic comparison: ängel before år
[edit] See also
transform a string so that strcmp would produce the same result as strcoll (function) |
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transform a wide string so that wcscmp would produce the same result as wcscoll (function) |