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您可以编写一种方法来完成所有这些操作，但是，它不会像高效的那样可读。您必须在通用或有效解决方案之间做出选择。
public static void main(String... args) throws IOException {
int[] nums = new int[10*1000 * 1000];
{
long start = System.nanoTime();
product2(nums);
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to take the product of %,d ints using an int[].%n", time / 1e9, nums.length);
}
{
long start = System.nanoTime();
product(nums);
long time = System.nanoTime() - start;
System.out.printf("Took %.3f seconds to take the product of %,d ints using reflections.%n", time / 1e9, nums.length);
}
}
public static double product(Object array) {
double product = 1;
for (int i = 0, n = Array.getLength(array); i < n; i++)
product *= ((Number) Array.get(array, i)).doubleValue();
return product;
}
public static double product2(int... nums) {
double product = 1;
for (int i = 0, n = nums.length; i < n; i++)
product *= nums[i];
return product;
}版画
Took 0.016 seconds to take the product of 10,000,000 ints using an int[].
Took 0.849 seconds to take the product of 10,000,000 ints using reflections.如果您只处理相对较小的阵列，那么通用但效率较低的解决方案可能足够快。