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🐛 Cannot call non-constexpr function in constexpr context (#2010)
Problem: - gcc-8 gives the following error when compiling this function on all standards: test/std-format-test.cc: In member function 'constexpr auto std::formatter<S>::parse(std::format_parse_context&)': test/std-format-test.cc:112:17: error: call to non-'constexpr' function 'int isdigit(int)' if (!isdigit(c) || (++iter, get_char()) != '}') ~~~~~~~^~~ Solution: - Write a `constexpr` version of `isdigit` for use in this function. Co-authored-by: Jonathan Gopel <jgopel@quantlab.com>
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@ -103,13 +103,17 @@ template <> struct std::formatter<S> {
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// Parses a width argument id in the format { <digit> }.
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constexpr auto parse(format_parse_context& ctx) {
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constexpr auto is_ascii_digit = [](const char c) {
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return c >= '0' && c <= '9';
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};
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auto iter = ctx.begin();
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// auto get_char = [&]() { return iter != ctx.end() ? *iter : 0; };
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auto get_char = [&]() { return iter != ctx.end() ? *iter : '\0'; };
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if (get_char() != '{') return iter;
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++iter;
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char c = get_char();
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if (!isdigit(c) || (++iter, get_char()) != '}')
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if (!is_ascii_digit(c) || (++iter, get_char()) != '}')
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throw format_error("invalid format");
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width_arg_id = fmt::detail::to_unsigned(c - '0');
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ctx.check_arg_id(width_arg_id);
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