Why does it skip array index 0?

Question

I came across this computer science problem and it wasn't working out the way I was writing it down. This is the code:

int[][]grid = {{1,2,3,4},{5,6,7},{8,9},{10}};

    for(int i = 0; i < grid.length; i++)
        for(int j = 0; j < grid[i].length; j++)
            grid[j][i] = grid[i][j];
    System.out.println(Arrays.toString(grid[1]));

It should change grid[0] to {1, 5, 8, 10} but instead it does nothing to it. Why does it skip over that one? Shouldn't i start out as 0 so the second for loop should start with grid[0][0] = grid[0][0] then grid[1][0] = grid[0][1]?

To change grid[0] to {1,5,8,10} you should do grid[i][j] = grid[j][i]; — Nadir
– Nadir, Commented Apr 22, 2016 at 8:22

Ben Green · Accepted Answer · 2016-04-22 08:44:10Z

3

It's because you're changing the initial variable (grid) on each iteration, put the output into a separate variable and then print that.

Explanation:

grid = {{1,2,3,4},{5,6,7},{8,9},{10}};

After the first sub loop (looping j)

grid = {{1,2,3,4},{2,6,7},{3,9},{4}};

Then when it performs the subsequent i loops, you can see it puts the numbers back where they were. If you start with an empty array as your output variable, you will avoid this problem.

edited Apr 22, 2016 at 8:44

answered Apr 22, 2016 at 8:34

Ben Green

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1 Comment

Ben Green Over a year ago

obviously if required for any reason, after the processing of the loops is complete, you can assign the output variable content back into the input variable (grid)

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