Doubt in Vectors

[techie929]

Hi,

I have a doubt regarding a function myfunc(int i,int j) function. The return value of this function is a vector of type int[ ] .For Example if we pass myfunc(4, 2) it will return a vector {( 1,0) ,( 0,1) ,( 1,0) ,( 1,0) , ( 0,1 ), ( 1,0)}.

Can anyone suggest how can I access the individual values.

public static Vector<int[]> myfunc(int i,int j){

	             Vector<int[]> result=new Vector<int[]>();
		int signOfI;
		int signOfJ;
		double ii=new Double(Math.abs(i));
		double jj=new Double(Math.abs(j));
		if(i==0 && j==0)
		return result;
		if(i>0) signOfI=1; else signOfI=-1;
		if(j>0) signOfJ=1; else signOfJ=-1;
					if (j==0){
					for (int f=0;f<ii;f++){
					int[] point={signOfI,0};
					result.add(point);
					}
					return result;
				   }
				if (i==0){
				for (int f=0;f<jj;f++){
				int[] point={0,signOfJ};
				result.add(point);
				}
				return result;
				}
				Double a=1.0;
				Double b=1.0;
				double cross=((ii/jj)*(a-.5)+.5);
				while (b<=ii || a<=jj){
				if (b>cross){
				int[] point={0,signOfJ};
				result.add(point);
				a++;
				cross=((ii/jj)*(a-.5)+.5);
				}else if (b==cross){
				int[] point={signOfI,signOfJ};
				result.add(point);
				a++;
				b++;
				cross=((ii/jj)*(a-.5)+.5);
				}else if (b<cross){
				int[] point={signOfI,0};
				result.add(point);
				b++;
				}
				}
				return result;
						}

I am trying to call the above function using

Vector<int[]> resultq=new Vector<int[]>();
    
    resultq=myFunc(4,2);
     for(int i=0; i<resultq.size(); i++)
     {
      System.out.println(resultq.get(i));
     }

But using this I am not able to get correct values.

[JamesCherrill]

When you get an element of that vector, you are getting an array of ints, which prints as an incomprehensible object reference. To print those in a readable way try the arrays.toString method to format the content of each array, eg

int[] arr = resultq.get(i);
System.out.println(Arrays.toString(arr));

System.out.println(Arrays.toString(resultq.get(i)));

[techie929]

Thanks a lot.