我想将一个从0到1的进度Double值转换为一个格式良好的String，其中包含选定数量的小数位，例如：0.789123456至“79.1%”
增加挑战：我想使用新的FormatStyle协议方法. formed（）。
令我惊讶的是，我不能让它工作。我找不到一种方法来指定精度。

var progress: Double = 0.789123456
progress.formatted(.percent) // "78,9123456 %" it starts so easy
progress.formatted(.number.precision(.fractionLength(1))) // "0,79"
progress.formatted(.number.precision(.fractionLength(0...1))) // "0,79"
progress.formatted(.number.precision(.fractionLength(0...1))).formatted(.percent) // does not compile
// "Instance method 'formatted' requires the types 'FloatingPointFormatStyle<Double>.FormatOutput' (aka 'String') and 'FloatingPointFormatStyle<Double>.Percent.FormatInput' (aka 'Double') be equivalent"

Double(progress.formatted(.number.precision(.fractionLength(1)))) // nil
Double("0.79")?.formatted(.percent) // "79 %", gotcha, I have german Locale!

Locale.current // "en_DE (current)"
Locale.current.decimalSeparator // ","

Double(progress.formatted(.number.precision(.fractionLength(1))).replacingOccurrences(of: Locale.current.decimalSeparator!, with: ".")) // "0,8"
Double(progress.formatted(.number.precision(.fractionLength(1))).replacingOccurrences(of: Locale.current.decimalSeparator!, with: "."))?.formatted(.percent) // "80 %"

有没有办法用FormatStyle来实现？或者我必须使用旧的NumberFormatter和它的maximumFractionDigits？