針對double類型的計算的問題之一：

public class DoubleCaculator {
    double price = 0.38;
    int count = 3;
    
    @Test
    public void demo1() {
        //結果：1.1400000000000001
        System.out.println(price * count);
    }
}

    @Test
    public void demo2() {
        //bigDecimal的字符串的構造完美解決了double計算的問題
        BigDecimal bdP = new
 
 BigDecimal(price);
        BigDecimal bdc = new BigDecimal(count);
        BigDecimal add = bdP.multiply(bdc);
        System.out.println(add.doubleValue());
    }

此結果是：2.2800000000000002

從上面兩個的結果，顯然與我們想要的結果有點不一樣，如果是針對金融方面的話，那麽一個小數點的問題就會影響很大，那麽對於這樣的問題，我們要如何的解決哪？

答:BigDecimal這個類。

那為啥上面使用了這個BigDecimal為啥還會出現這個問題？

答：傳入的參數有問題。

源碼的註釋說明：如果B中傳入的為double類型的值時的解釋

The results of this constructor can be somewhat unpredictable.
     * One might assume that writing {@code new BigDecimal(0.1)} in
     * Java creates a {@code BigDecimal} which is exactly equal to
     * 0.1 (an unscaled value of 1, with a scale of 1), but it is
     
 
* actually equal to
     * 0.1000000000000000055511151231257827021181583404541015625.
     * This is because 0.1 cannot be represented exactly as a
     * {@code double} (or, for that matter, as a binary fraction of
     * any finite length).  Thus, the value that is being passed
     * <i>in</i> to the constructor is not exactly equal to 0.1,
     * appearances notwithstanding.
     *

以及給出的最佳的解決方法：

The {@code String} constructor, on the other hand, is
     * perfectly predictable: writing {@code new BigDecimal("0.1")}
     * creates a {@code BigDecimal} which is <i>exactly</i> equal to
     * 0.1, as one would expect.  Therefore, it is generally
     * recommended that the {@linkplain #BigDecimal(String)
     * <tt>String</tt> constructor} be used in preference to this one.
     *

所以上面的代碼如果將傳入的double類型的值轉換成字符串的話，那麽問題就解決了：

    @Test
    public void demo2() {
        //bigDecimal的字符串的構造完美解決了double計算的問題
        BigDecimal bdP = new BigDecimal(price + "");
        BigDecimal bdc = new BigDecimal(count + "");
        BigDecimal add = bdP.multiply(bdc);
        System.out.println(add.doubleValue());
    }

解決double的值相加的問題