How to avoid repeated code in java?

Question

I'm just starting to program. uf is a join-search class with a union of methods that binds the roots of two nodes. This piece of code is responsible for opening the grid site and combining the site with its neighbor, if any of the neighbors opens. And if one of its neighbors is full, fill in all the nodes associated with the site. This is the actual code:

if(i == 1){ uf.union(len*len, xyTo1D(i,j)); if(existAndOpen(i+1,j)){ uf2.union(xyTo1D(i+1,j), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i+1,j)); } if(existAndOpen(i-1,j)){ uf2.union(xyTo1D(i-1,j), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i-1,j)); } if(existAndOpen(i,j-1)){ uf2.union(xyTo1D(i,j-1), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i,j-1)); } if(!(j == len && i == len)){ if(existAndOpen(i,j+1)){ uf2.union(xyTo1D(i,j+1), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i,j+1)); } } } else{ if(existAndFull(i+1,j)){ uf2.union(xyTo1D(i+1,j), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i+1,j)); } if(existAndFull(i-1,j)){ uf2.union(xyTo1D(i-1,j), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i-1,j)); } if(existAndFull(i,j-1)){ uf2.union(xyTo1D(i,j-1), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i,j-1)); } if(!(j== len && i == len)){ if(existAndFull(i,j+1)){ uf2.union(xyTo1D(i,j+1), xyTo1D(i,j)); uf.union(xyTo1D(i,j), xyTo1D(i,j+1)); } } if(existAndOpen(i+1,j)){ uf.union(xyTo1D(i,j), xyTo1D(i+1,j)); } if(existAndOpen(i-1,j)){ uf.union(xyTo1D(i,j), xyTo1D(i-1,j)); } if(existAndOpen(i,j-1)){ uf.union(xyTo1D(i,j), xyTo1D(i,j-1)); } if(!(j== len && i == len)){ if(existAndOpen(i,j+1)){ uf.union(xyTo1D(i,j), xyTo1D(i,j+1)); } } } } 

How can I simplify the code?

Answer 1

try it

 boolean f1(int a, int b) { } boolean f2(int a, int b) { } void A(int a, int b) { } void testAndA(BiPredicate<Integer, Integer> p, int a, int b) { if (p.test(a, b)) A(a, b); } 

and

  if(x == 1){ testAndA(this::f1, x + 1, y); testAndA(this::f1, x, y + 1); } else { testAndA(this::f2, x + 1, y); testAndA(this::f2, x, y + 1); } 

Answer 2

You can write a loop to see all the different values ​​that can be passed to f1 (), for example:

 for (int deltax = -1; deltax <= 1; deltax++) { for (int deltay = -1; deltay <= 1; deltay++) { if (f1(x + deltax, y + deltay)) { A(x + deltax, y + deltay); } } } 

Of course, change the start and end values ​​of deltax and deltay depending on what conditions you need to check.

Answer 3

You say that in each block there are "more if statements". Assuming that this means you need to make more calls to f1 / f2 and A , but with different offsets x and y .

Here is a program that shows a way to reorganize code to avoid repeating coding. Its main functions are:

• It uses object orientation to abstract the internal if blocks.
• It uses the offsets array to represent the x / y offsets for each inner if block.

• The doIt() method uses a loop to call if blocks inside.

   public class Main { static interface F { void f(int i, int j); } static class F1Caller implements F { public void f(int a, int b) { if (f1(a, b)) { A(a, b); } } } static class F2Caller implements F { public void f(int a, int b) { if (f2(a, b)) { A(a, b); } } } static boolean f1(int a, int b) { System.out.print(" f1. "); return true; } static boolean f2(int a, int b) { System.out.print(" f2. "); return true; } static void A(int a, int b) { System.out.println("a: " + a + ", b: " + b); } static F1Caller f1Caller = new F1Caller(); static F2Caller f2Caller = new F2Caller(); // x and y offsets for each call to f1/f2. // Add more offset rows, as needed. static int offsets[][] = { {1, 0}, {0, -1} }; static void doIt(int x, int y) { System.out.println("x: " + x + ", y: " + y); F f = (x == 1) ? f1Caller : f2Caller; for (int k = 0; k < offsets.length; k++) { ff(x + offsets[k][0], y + offsets[k][1]); } } public static void main(String[] args) { doIt(0, 0); doIt(1, 0); } } 

The output of the specified program:

 x: 0, y: 0 f2. a: 1, b: 0 f2. a: 0, b: -1 x: 1, y: 0 f1. a: 2, b: 0 f1. a: 1, b: -1 

Answer 4

You can combine the function together.

Using the switch.

 public static boolean func(int a, int b, int fun) { boolean output = false; switch(fun) { case 1: //do stuff output = true; break; case 2: //do stuff output = true; break; default: //unknown function handling output = false; } return output; } 

Integrate it with a for loop:

 public static void main(String[] args) { int i; //Change this to whatever you want or set it to a argument. int repeat = 2; for(i = 1; i <= repeat; i++) { func(a, b, i); } }