remove element without causing ConcurrentModificationException & i18n number input solution & 免費錄屏軟件
1. remove element without causing ConcurrentModificationException
List<String> names =
.
Iterator<String> i = names.iterator();
while (i.hasNext()) {
// Do something
i.remove();
}

Iterator<String> i = names.iterator();
while (i.hasNext()) {
// Do something
i.remove();
}
Note that you must call i.next() before you can call i.remove()
2. 法語,德語中數(shù)字小數(shù)點用,表示,千位符是"." 如果做支持I18n的UI, 需要顯示數(shù)字并接受他們輸入的數(shù)字,可以這樣做:
3. 免費的錄屏軟件:HyperCam
2. 法語,德語中數(shù)字小數(shù)點用,表示,千位符是"." 如果做支持I18n的UI, 需要顯示數(shù)字并接受他們輸入的數(shù)字,可以這樣做:
public static void main(String[] args) throws ParseException{
String numString = "0.05";
String frenchNumString = "0,05";
NumberFormat nf = NumberFormat.getNumberInstance();
System.out.println("French format of raw string: " + nf.format(Double.parseDouble(numString)));
Double n = (Double) nf.parse(frenchNumString);
System.out.println("Parse french number and calculate: " + (n + 1));
}
String numString = "0.05";
String frenchNumString = "0,05";
NumberFormat nf = NumberFormat.getNumberInstance();
System.out.println("French format of raw string: " + nf.format(Double.parseDouble(numString)));
Double n = (Double) nf.parse(frenchNumString);
System.out.println("Parse french number and calculate: " + (n + 1));
}
3. 免費的錄屏軟件:HyperCam