如何在object -c中交换两个数组

Is this a convenient way for swapping two ivar arrays in objective C?

在objective C中交换两个ivar数组是否方便?

- (void) foo {
    NSArray *aux;
    aux = array1;
    array2 = array1;
    array1 = array2;
}

Are there any alternatives? May it have problems related to retainCount under some circumstances? I am confused because in the program that I am reviewing the swap is done by:

有什么选择吗?在某些情况下，会不会有与遗产计算有关的问题?我很困惑，因为在我正在审查的项目中，是由:

- (void) foo {
    NSArray *aux;
    aux = array1;
    [aux retain];
    array2 = array1;
    array1 = array2;
    [aux release];
}

3 个解决方案

#1

In your example, you're swapping just the pointers. And by the way, you're doing it wrong:

在你的例子中，你只交换指针。顺便说一下，你做错了:

- (void) foo {
    NSArray *aux;
    aux = array1;       /* aux is array1 */
    array2 = array1;    /* array2 is array1 */
    array1 = array2;    /* array1 is array1 */
}

If you have properties on those arrays, I'd recommend going with it:

如果你在这些数组上有属性，我建议你这样做:

- (void)swapMyArrays
{
    NSArray *temp = [NSArray arrayWithArray:self.array1];
    [self setArray2:array1];
    [self setArray1:temp];
}

#2

Neither of them will work, you are not using your aux variable anywhere, at the end they will both point to the same memory location do this instead:

它们都不能工作，你没有在任何地方使用你的辅助变量，最后它们都指向相同的内存位置，这样做:

- (void) foo {
    NSArray *aux;
    aux = array1;
    array1 = array2;
    array2 = aux;
}

#3

I think you have a mistake at the end, and it should be:

我认为你最后犯了一个错误，应该是:

- (void) foo {
    NSArray *aux;
    aux = array1;
    array1 = array2;
    array2 = aux; // instead of array2 = array1
}

otherwise you make both arrays the same one.

否则，将使两个数组相同。

I edited the answer as for the (correct) comments regarding a mistake in the variables I made.

我编辑了关于我所犯的变量错误的(正确的)评论的答案。

#1