The calculation of large numerical value in Java

Source: Internet
Author: User

I believe that everyone may encounter to the value of more than 17-bit integer, Java in the value of double is more than 8 bits will become a scientific notation, so I used the DecimalFormat format, when the pattern is "#################.##", The value of "123456789012345.78" can be displayed normally, not formatted with 1.2345678901234578E14 formatted 123456789012345.78 but the value of more than 15-bit integer case is not the same, such as the value of " 1234567890123456.78 "After the format is 1234567890123456.8, has been rounded up." Finally checked Javadoc, found in the Java.math package has a class of bigdecimal can solve the problem of large numerical calculation, is quite good, and finally solved the problem.

Example://Test BigDecimal

BigDecimal bd=new BigDecimal ("12345678901234578.789");

SYSTEM.OUT.PRINTLN ("BigDecimal * 12.5=" + (bd=bd.multiply (new BigDecimal (12.5)));

System.out.println ("bigdecimal/12.5=" +bd.divide (New BigDecimal (12.5));

Output is: BigDecimal * 12.5=154320986265432234.8625


Absolutely right!

Attention: BigDecimal when used to pay attention to it in the form of internal representation of the computer, the new BigDecimal (double) This constructor generated BigDecimal class will become unpredictable, because the 0.100000000******* To represent 0.1, and use 0.29999999***** to indicate that if you do not process the results, there may be bugs.

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