我有此代码:
while (counter <= t) { a = a * counter; c = c * r + a; counter++; },我有 t ,其值为 101835 .
输出应为7.1 * 10 ^ 438,但NetBeans显示输出为 infinity .
The output should be 7.1*10^438, but NetBeans shows the output as infinity.
有什么东西可以将输出显示为十进制数字?
Is there anything that will make the output as a decimal number?
推荐答案A double 不能代表这样的数字,它太大了.范围大约在-10 308 和10 308 之间.
A double as defined by IEEE-754 can't represent such a number, it's too large. The bounds are approximately between -10308 and 10308.
您需要使用 BigDecimal 来表示它:一个具有任意字节数的数字来表示数字.
You need to use a BigDecimal to represent it: a number with an arbitrary number of bytes to represent numbers.
实现此目标的更好方式:
Better way to implement this:
double c = 0.0005d;//initialize c double r = 0.01d; //initialize r double a = 0.0006d;//initialize a BigDecimal abd = new BigDecimal(a); //BigDecimal for a BigDecimal cbd = new BigDecimal(c); //BigDecimal for c BigDecimal rbd = new BigDecimal(r); //BigDecimal for r for (int counter = 1; counter <= t; counter++) {//perhaps other offset for counter? abd = abd.multiply(new BigDecimal(counter)); cbd = cbd.multiply(rbd).add(abd); }这种方法的潜在问题是精度太高:Java将精确地计算所有操作,从而得出具有数千位数字的数字.由于每个运算都会炸掉位数,因此在几次迭代中,简单的加法和乘法运算变得不可行.
A potential problem with this approach is that the precision is too high: Java will calculate all operations exactly resulting in numbers that have thousands of digits. Since every operation blows up the number of digits, within a few iterations, simple addition and multiplication operations become unfeasible.
您可以通过使用可选的 MathContext 参数定义精度来解决此问题:它确定结果的精度.例如,您可以使用 MathContext.DECIMAL128 :
You can solve this by defining a precision using the optional MathContext parameter: it determines on how precise the result should be. You can for instance use MathContext.DECIMAL128:
int t = 101835; double c = 0.0005d;//initialize c double r = 0.01d; //initialize r double a = 0.0006d;//initialize a BigDecimal abd = new BigDecimal(a); //BigDecimal for a BigDecimal cbd = new BigDecimal(c); //BigDecimal for c BigDecimal rbd = new BigDecimal(r); //BigDecimal for r for (int counter = 1; counter <= t; counter++) {//perhaps other offset for counter? abd = abd.multiply(new BigDecimal(counter),MathContext.DECIMAL128); cbd = cbd.multiply(rbd,MathContext.DECIMAL128).add(abd,MathContext.DECIMAL128); } System.out.println(abd); System.out.println(cbd);这给出了:
abd = 3.166049846031012773846494375835059E+465752 cbd = 3.166050156931013454758413539958330E+465752这大概是正确的,毕竟 a 的结果应该是:
This is approximately correct, after all the result of a should be:
根据 Wolfram Alpha ,这大约是正确的.
Which is approximately correct according to Wolfram Alpha.
此外,如果是 for 循环,我建议使用 for ,而不要使用 while .由于 while 倾向于创建另一种类型的无穷大:无限循环;).
Furthermore I would advice to use a for and not a while if it is a for loop. Since while tends to create another type of infinity: an infinite loop ;).
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