java - 最小位程序

Question

我是 Java 新手。我尝试从 Java Just in Time 中学习它。

作者:

So program to calculate how many bits would be needed to represent a given number of different values.

程序是:

int numberOfValues = Integer.parseInt(args[0]);
int noOfBits = 0;

while (Math.pow(2, noOfBits) < numberOfValues)
  noOfBits = noOfBits + 1;

但是书作者被要求用其他版本写这个。

作者:

In this task you will write a variation of the MinimumBitWidth program which works a little more efficiently. Instead of computing a power of 2 in the loop condition on each iteration, your version will accumulate 2 to the power of noOfBits in a separate variable. This can be done by initialising your new variable to 1, and simply doubling its value each time you increment noOfBits.

You will use the same test data as used for the previous version of the program.

可能是因为英语不好，我听不懂他的意思。感谢帮助。 :)

Answer 1

意思是将数字与 2⁰, 2¹, ... 2³¹ 进行比较。由于 2³¹ 大于 Integer.MAX_VALUE，因此可能会使用 long。

int noOfBits = 0;
long pow2 = 1;
while (numberOfValues >= pow2) {
    ++noOfBits;
    //pow2 = 2 * pow2
    //pow2 += pow2;
    //pow2 *= 2;
    pow2 <<= 1; // Shift left once.
}

另见 Integer.numberOfLeadingZeros(int) ，这会将代码减少为一行代码。

int noOfBits = 32 - Integer.numberOfLeadingZeros(numberOfValues);