"Java doubts" Long integer calculation prevents overflow

The following code:

public class Example003 {public static void main (string[] args) {final long Micros_per_day = $ * * * + * 1000;FI nal Long MICROS_PER_DAY_L1 = 24 * * * * * * * 1000l;final Long millis_per_day = 60 * 60 * 1000; System.out.println ("NO L Print:" + micros_per_day/millis_per_day); Output 1system.out.println ("has L Print:" +micros_per_day_l1/millis_per_day); Output 2}}

Output Result:

NO l Print:5has L print:1000

Cause Analysis:

The results of output 1 and 2 are inconsistent because of the calculation overflow. Since all the multiplying factors in micros_per_day are int, when the two int is multiplied, the result is an int, so the micros_per_day evaluates to int, which is then assigned to the long type after the calculation is complete (overflow) micros_per_ Day,micros_per_day gets a partial value after overflow; MICROS_PER_DAY_L1 is calculated as a long type, and the result of the calculation is a long, which does not produce overflow. Therefore, in the operation of large numbers, must be aware of the overflow problem. Typically, a long is used to execute when calculating. That is, the standard numeric type is explicitly in the calculated factor.

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"Java doubts" Long integer calculation prevents overflow