io: remove the view
overload for slurp
This just confuses matters. A `view` can be across arbitrary data, and is often used as a representation of a file's _data_.
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6
io.hpp
6
io.hpp
@ -31,12 +31,6 @@ namespace cruft {
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template <typename T = std::byte>
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template <typename T = std::byte>
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std::vector<T> slurp (const std::experimental::filesystem::path&);
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std::vector<T> slurp (const std::experimental::filesystem::path&);
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template <typename T = std::byte, typename A, typename B>
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std::vector<T> slurp (cruft::view<A,B> path)
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{
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const std::experimental::filesystem::path src { path.begin (), path.end () };
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return slurp<T> (src);
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}
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template <typename T = std::byte>
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template <typename T = std::byte>
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std::vector<T> slurp (FILE *);
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std::vector<T> slurp (FILE *);
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