使用 Double.parseDouble() 时如何禁用自动转换为指数表达式？

Question

When I run the following program:

System.out.println(Double.parseDouble("99999999999999") + 1);

I got: 1.0E14

How can I disable the auto-conversion so that I get 100000000000000.0 ?

Answer 1

That has nothing to do with Double.parseDouble , and everything to do with Double.toString .

Try this:

System.out.printf("%.1f", Double.parseDouble("99999999999999") + 1);

Answer 2

Use aBigDecimal instead.

    import java.math.BigDecimal;

    BigDecimal big = new BigDecimal("99999999999999");
    big = big.add(BigDecimal.ONE); // Thanks @Chris for this tip
    System.out.println(big); // Prints "100000000000000"

BigNumbers offer arbitrary precision. Double is limited to an (approximated) 64-bit representation. If you were to use too many 9's, you would get the wrong answer:

    System.out.printf("%.1f", Double.parseDouble("9999999999999998") + 1); // Note ends in 8

This prints "10000000000000000.0" , which is incorrect, should be "9999999999999999.0" , but the number has exceeded the accuracy of Double