得到了两种方法，用相同的数组，用sane公式转换数组元素。方法返回不同的值

Question

private static final int SIZE = 10000000;
private static final int h = SIZE/2;

  private static float[] method1() {
    float arr[] = new float[SIZE];

        //array fill with 1,                                                        
    for (int i = 0; i < arr.length; i++) {
        arr[i] = 1;
    }


       //calulate new values with new formula                                                              
     for (int i = 0; i < arr.length; i++) {
        arr[i] = (float) (i * Math.sin(0.2f + i / 5) * 
                              Math.cos(0.2f + i / 5) * 
        Math.cos(0.4f + i / 2));

    }
     //return new value
        return arr;
}

 private static float[] method2(){
    float arr[] = new float[SIZE];

    float firstarr[] = new float[h];

    float secondarr[] = new float[h];

    float result[] = new float[SIZE];

    for (int i = 0; i <arr.length ; i++) {
        arr[i] = 1;
    }
      //Devide array in two with same length arr/2,                                                                            
    System.arraycopy(arr, 0, firstarr, 0, h);
    System.arraycopy(arr, h, secondarr, 0, h);


    for (int i = 0; i < firstarr.length; i++) {
        firstarr[i] = (float) (i * Math.sin(0.2f + i / 5) *      
                                   Math.cos(0.2f + i / 5) *      
                                   Math.cos(0.4f + i / 2));
    }

    for (int i = 0; i < secondarr.length; i++) {
        secondarr[i] = (float) (i * Math.sin(0.2f + i / 5)                   
                                  * Math.cos(0.2f + i / 5)                   
                                  * Math.cos(0.4f + i / 2));
    }
    //compile back in one                                                                                

    System.arraycopy(firstarr, 0, result, 0, h);
    System.arraycopy(secondarr, 0, result, h, h);

    // return result
    return result;

}

    private static long amount(float[] arr){
    long result = 0;

    for (int i = 0; i <arr.length ; i++) {
       result += arr[i];
    }
    return result;
}




public static void main(String[] args) {
    System.out.println(amount(method1()));// ammount for first method: 22527562

    System.out.println(amount(method2())); // ammount for second method: -20047478

}}

Answer 1

你的问题是循环。第一个方法循环i从0到N，第二个方法循环两次从i到N/2。

例如：对于N=4，第一个方法返回您

i=0 0.0 
i=1 0.17....
i=2 0.066....
i=3 0.099....

但是第二个：

i=0 0.0 
i=1 0.17....
i=0 0.0 
i=1 0.17....

    for (int i = h; i < SIZE; i++) {
    secondarr[i-h] = (float) (i * Math.sin(0.2f + i / 5)                   
                              * Math.cos(0.2f + i / 5)                   
                              * Math.cos(0.4f + i / 2));
}