Double.parseDouble（“100d”）如何没有错误

Question

Double.parseDouble("100a"); // getting the error as expected But why does the below line doesn't gets the error

Double.parseDouble("100d");

Please explain me.

Answer 1

According to the source code, the Double.parseDouble(String s) method uses the FloatingDecimal.readJavaFormatString(String s) to parse the provided argument.

There, you can see that there's a check if the provided String ends with 'D' , 'd' , 'F' or 'f' and if it doesn't, then a NumberFormatException is thrown:

if (i < l &&
   ((i != l - 1) ||
    (in.charAt(i) != 'f' &&
     in.charAt(i) != 'F' &&
     in.charAt(i) != 'd' &&
     in.charAt(i) != 'D'))) {
       break parseNumber; // go throw exception
     }

Answer 2

d represents double .

public static double parseDouble(String s) throws NumberFormatException

Throws:
NumberFormatException - if the string does not contain a parsable double.

100d is a parsable double string that's why it doesn't show any error.

Answer 3

The javadoc says:

Returns a new double initialized to the value represented by the specified String, as performed by the valueOf method of class Double.

So we look at the javadoc for Double::valueOf which gives the detailed acceptable syntax, in particular:

Sign _opt FloatingPointLiteral

where FloatingPointLiteral is as defined in the JLS. So we go to JLS #3.10.2 and you see that it can end with f , F , d or D , as expected.

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If you want to only accept "human-readable doubles", you can use a DecimalFormat .