Correct me if i'm wrong, but doesnt integer stop at like... 2.7bil?
And if i look at:
And make it easier (for me) to read: 9'223'372'030'926'249'001, which are (rounded up) 10 billion billions, or about 400 times the max of a 'reliable' integer value.
And as on overflow, they stop counting at the ~2.7bils
(overnighted and not used to such high numbers, might be mathematical incorrect proportions, but you get the idea)
One would need (afaik) a long int or int64, which are both (afaik) not available to the bash shell.
hth
So much for getting some sleep...
According to the standards, conforming shells must use signed long int (or something else that provides the same results when working with values in the range of a signed long int). With the bash being talked about on this thread:
shows the results required for a conforming shell in the 1st two examples and shows an allowed (and expected) result for the 3rd case where the last operand is out of the range of a long int (signed or unsigned).
I agree that it is hard to count the digits when looking at numbers that big, but the shell doesn't allow thousands separators in integer constants given to it in expressions nor when they are used as operands for shell built-ins expecting integer arguments.
Last edited by Don Cragun; 08-21-2015 at 01:03 PM..
Reason: Add note about large integer constants in the shell.
This User Gave Thanks to Don Cragun For This Post:
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