英文:
Create BigDecimal from unscaled long
问题
我正在尝试将长整数1099转换为BigDecimal的10.99；这给我了11.00：long cost = 1099; MathContext CENTS = new MathContext(2,RoundingMode.HALF_EVEN); BigDecimal result = (new BigDecimal(cost,CENTS)).movePointLeft(2); 据我所知，这应该可以工作。我的愚蠢错误是什么？
英文:
I'm trying to convert long 1099 to BigDecimal 10.99;
This gives me 11.00:
long cost = 1099;
MathContext CENTS = new MathContext(2,RoundingMode.HALF_EVEN);
BigDecimal result = (new BigDecimal(cost,CENTS)).movePointLeft(2);
AFAIK this should work. What's my bonehead error?
答案1
得分: 2
错误在于精度和标度之间有一个区别。 MathContext 的构造函数接受一个精度，这是小数点两边的十进制数字的总数。 （例如，您最初的 BigDecimal 实际上是 11 * 10^2，就好像它是科学计数法。）
将其更改为 new MathContext(4, RoundingMode.HALF_EVEN)。
英文:
The error is that there's a distinction between scale and precision. The constructor of MathContext accepts a precision, which is a total number of decimal digits on either side of the decimal point. (For example, the original BigDecimal you had was essentially 11 * 10^2, as if it were in scientific notation.)
Change it to new MathContext(4, RoundingMode.HALF_EVEN).
答案2
得分: 0
使用仅接受 long 参数的构造函数，不需要 MathContext。
英文:
new BigDecimal( 1099L )
.movePointLeft(2)
Use constructor taking only a long, without the MathContext.
See this code run live at IdeOne.com.
答案3
得分: 0
Use the BigDecimal.valueOf(long unscaledVal, int scale) method:
long cost = 1099;
BigDecimal result = BigDecimal.valueOf(cost, 2);
System.out.println(result); // 10.99
英文:
Use the BigDecimal.valueOf(long unscaledVal, int scale) method:
long cost = 1099;
BigDecimal result = BigDecimal.valueOf(cost, 2);
System.out.println(result); // 10.99
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