为什么Long.parseLong（String s）不会将尾随的“L”视为有效？

Question

Long myLong = Long.parseLong("1L") // Throws NumberFormatException

My best guess is that an additional check like that would cause an overhead which would only come into play ~1% (or less) of the time anyone calls parseLong with a trailing "L" character.

But is there perhaps a different reason behind this "omission"?

Answer 1

It is defined so in the method java.lang.Long.parseLong(String):

The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign ...

Answer 2

The trailing L (or l ) is only relevant to the Java compiler to distinguish int literals from long literals. It's not a universal marker for long values, so accepting (and ignoring) it would be more of a weird side effect, not to mention a source for bugs.

Answer 3

Thinking about the use case of Long.parseLong("1L"); is you got long value in the form of String with l at the end.

Now how can you get this value in real life applications, consider GUI or console application, conversion of number String to actual type will not have suffix l or d . From user input to database value suffix l is not acceptable for number input. You will never get String value as 1L from the business logic for the conversion to actual Long value.

Surprisingly, on the other hand Double myDouble = Double.parseDouble("1d"); will work fine because sun.misc.FloatingDecimal managed to parse with suffix d in number because double value may contain other characters like E or e in it for exponential term.

Answer 4

The L in 1L indicates to the Java compiler that the value is a long literal . It only has meaning in the Java source code for disambiguation with 1 (an int literal).

When you use Long.parseLong(String) with the string "1L" , it is a runtime error, because L is not a valid numerical digit in base 10. There is also no need for disambiguation, as you have already explicitly declared you want a Long as the result.

Or more formally, the documentation of Long.parseLong(String) specifies:

Parses the string argument as a signed decimal long. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign ' - ' ( \- ') to indicate a negative value or an ASCII plus sign ' + ' (' \+ ') to indicate a positive value. The resulting long value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseLong(java.lang.String, int) method.

Note that neither the character L (' \L ') nor l (' \l ') is permitted to appear at the end of the string as a type indicator, as would be permitted in Java programming language source code.

This last paragraph explicitly documents it is not allowed.