开发者

Why all elements of my list seems to be the same?

开发者 https://www.devze.com 2023-01-28 00:22 出处:网络
I have the following code: Integer[] lastExchange = new Integer[nColors]; Integer[] newExchange = new Integer[nColors];

I have the following code:

Integer[] lastExchange = new Integer[nColors];
Integer[] newExchange = new Integer[nColors];
while (true) {
    ...
    for (int i=0; i<nColors; i++) {
       lastExchange[i] = newExchange[i];
    }
    ...
    exchanges.add(lastExchan开发者_如何转开发ge);
    output.log.fine("Exchange:" + lastExchange[0] + "," + lastExchange[1]);
}
for (Integer[] exchange : exchanges) {
    output.log.fine("Exchange:" + exchange[0] + "," + exchange[1]);
}

I have two outputs (one in the while loop another one in the for loop). The first output shows me that I do add different arrays to the list. While when i do a double check in the second loop I see that all elements of the exchange list are the same (they are equal to the first element of the list).

Does anybody know what I am doing wrong here?


As unwind's answer states, you're adding a reference to the same array in every iteration of the loop. You need to create a new array each time:

// It's not clear where newExchange is actually populated
Integer[] newExchange = new Integer[nColors];
while (true) {
    Integer[] lastExchange = new Integer[nColors];
    ...
    for (int i=0; i<nColors; i++) {
       lastExchange[i] = newExchange[i];
    }
    ...
    exchanges.add(lastExchange);
    output.log.fine("Exchange:" + lastExchange[0] + "," + lastExchange[1]);
}

Alternatively, if you're just cloning the array:

Integer[] newExchange = new Integer[nColors];
while (true) {
    Integer[] lastExchange = newExchange.clone();
    ...
    exchanges.add(lastExchange);
    output.log.fine("Exchange:" + lastExchange[0] + "," + lastExchange[1]);
}


What is the type of lastExchange? If it's a reference to an object, the the problem is probably exactly that; you just add the same reference to a mutable object, which is then modified and added again.

Since the first loop prints the object before it (presumably) is modified, it prints the proper (different) values.

0

精彩评论

暂无评论...
验证码 换一张
取 消