Java通过引用和编译器优化传递 [英] Java pass by reference and compiler optimization

问题描述

在函数中，fermatFactorization（）， a 和 b 被传递为引用参数，因为我正在使用 Long Class。但是，当我传递 a 和 b testFermatFactorization（） c>到> fermatFactorization（）， a 和 b 不要改变，所以 testFermatFactorization（）打印（0）（0）。我在 fermatFactorization（） a 和 b c>，我得到了我期望的输出。

我忽略了什么？编译器能否在 fermatFactorization（）中改变 a 和 b > 因为它们只被分配给？（可疑）

  public static void fermatFactorization（Long n，Long a，Long b）
 // PRE：n是要分解的整数
 // POST：a和b将是因子n 
 {
 Long v = 1L; 
 Long x =（（Double）Math.ceil（Math.sqrt（n）））。longValue（）; 
 //System.out.println(\"x:+ x）; 
 Long u = 2 * x + 1; 
 Long r = x * x  -  n; （r！= 0）//我们正在寻找条件x ^ 2  -  y ^ 2  -  n为零
 {
 while（r> 0） 
 {
r = r  -  v; //更新我们的条件
 v = v + 2; （r <0）
 {$ b $ //跟踪（y + 1）^ 2  -  y ^ 2 = 2y + 1，增加y
} 
 br = r + u; 
 u = u + 2; //跟踪（x + 1）^ 2  -  x ^ 2 = 2x + 1，增加x
} 
} 
 
a =（u + v  - 2）/ 2; //记住你和v相等; - > （2x + 1 + 2y + 1-2）/ 2 = x + y 
 b =（u-v）/ 2; //  - > （2x + 1-（2y + 1））/ 2 = x-y 
} 
 
 public static void testFermatFactorization（Long number）
 {
 Long a = 0L; 
长b = 0L; 
 fermatFactorization（number，a，b）; 
 System.out.printf（Fermat分解（％d）=（％d）（％d）\\\
，数字，a，b）; 
 
  

解决方案

Java是按值传递的。如果为参数分配新值，则不会影响调用方法中的值。

您有两个选择：
$ b

• 使您的方法返回 a 和 b - 可以在 int [] 中或使用一个单独的 FactorizationRezult 类有两个字段。这样你将在被调用的方法中声明 a 和 b 作为局部变量，而不是将它们作为参数。这是最可取的方法。

另一种方法是使用 MutableLong 并使用 setValue（..）方法 - 这种方式的变化将影响调用方法中的对象。这是不明智的

In the function fermatFactorization(), a and b are being passed as reference parameters, since I am using the Long Class. However, in function testFermatFactorization() when I pass a and b to fermatFactorization(), the values of a and b do not get changed, and so testFermatFactorization() prints (0)(0). I tested this by printing out a and b in fermatFactorization(), and I got the output that I expected.

What am I overlooking? Could the compiler alter a and b in fermatFactorization() since they are only being assigned to?(doubtful)

public static void fermatFactorization(Long n, Long a, Long b)   
//PRE:  n is the integer to be factored
//POST: a and b will be the factors of n
{
    Long v = 1L;
    Long x = ((Double)Math.ceil(Math.sqrt(n))).longValue();
    //System.out.println("x: " + x);
    Long u = 2*x + 1;
    Long r = x*x - n;

    while(r != 0)                 //we are looking for the condition x^2 - y^2 - n to be zero
    {
        while(r>0)
        {
            r = r - v;            //update our condition
            v = v + 2;            //v keeps track of (y+1)^2 - y^2 = 2y+1, increase the "y"
        }
        while(r<0)
        {
            r = r + u;
            u = u + 2;            //keeps track of (x+1)^2 - x^2 = 2x+1, increases the "x"
        }
    }

    a = (u + v - 2)/2;            //remember what u and v equal; --> (2x+1 + 2y+1 - 2)/2 = x+y
    b = (u - v)/2;                //                             --> (2x+1 -(2y+1))/2 = x-y
}

public static void testFermatFactorization(Long number)
{
    Long a = 0L;
    Long b = 0L;
    fermatFactorization(number, a, b);
    System.out.printf("Fermat Factorization(%d) = (%d)(%d)\n", number, a, b);
}

解决方案

Java is pass by value. If you assign a new value to the argument, it won't affect the value in the caller method.

You have two options:

• make your method return a and b - either in a int[] or using a separate FactorizationRezult class that has two fields. That way you will declare a and b as local variables in your called method, rather than taking them as parameters. This is the most advisable approach.

• An alternative approach is to use a MutableLong and use a setValue(..) method - that way the changes will affect the object in the caller method. This is less advisable

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