问题
This question already has an answer here:
- Why is 08 not a valid integer literal in Java? 6 answers
I am using quartz for schedulling.
TriggerUtils.getDateOf(0,40,18,09,06);
it accept 5 parameter. (seconds, minutes, hours, daysOfMonth, month).
When i pass fourth parameter as \"09\". Eclipse give me error \"The literal Octal 09 (digit 9) of type int is out of range \".
But when i pass the fourth parameter as \"9\" instead of \"09\", it works.
Can anyone explain me this error?
回答1:
In java, if you are defining an integer, a leading '0' will denote that you are defining a number in octal
int i = 07; //integer defined as octal
int i = 7; // integer defined as base 10
int i = 0x07; // integer defined as hex
int i = 0b0111; // integer defined as binary (Java 7+)
回答2:
There is no 9 in Octal (what you get with a preceding 0). 0-7, only.
回答3:
When you precede a number with 0 ("09" rather than "9"), then Java (and C and many other languages) interpret the number to be in octal - base-8.
"09" is not a valid number in octal - any single digit can be a maximum of "7" (since in octal, numbers go from 0..7).
回答4:
Numbers that begin with the zero digit are treated as octal (base 8) literals, and 9 is not a valid octal digit.
回答5:
10 is how many digits you have, whereas 010 is is what you get if you don't count your thumbs.
来源：https://stackoverflow.com/questions/970039/09-is-not-recognized-where-as-9-is-recognized