假设我想对二维数组进行排序。（只需重新排列行，不要触摸每行内的数据）。

在以下代码段中：所有 3 种情况都使用相同的Arrays.sort(T[] a, Comparator<? super T> c)方法签名。情况 (a) 工作正常。但是，只需在第二个参数中添加一个 if 条件，T 的推理就会发生变化。我无法理解为什么。

        // array contains 3 tuples, sort it by the first element, then second element
        int[][] array1 = new int[3][2];
        array1[0] = new int[]{1,2};
        array1[1] = new int[]{2,3};
        array1[2] = new int[]{2,4};

        // Case (a): compiles good, tuple is inferred as int[]
        Arrays.sort(array1, Comparator.comparingInt(tuple -> tuple[0]));  // Arrays.sort(T[] a, Comparator<? super T> c) correctly infers that T refers to int[]

        // Case (b.1): compile error: incompatible types
        // tuple is now inferred as Object, why?
        Arrays.sort(array1,
                (a1, a2) -> a1[0] == a2[0] ?
                        Comparator.comparingInt(tuple -> tuple[1]) : Comparator.comparingInt(tuple -> tuple[0]));  

        // Case (b.2): compile error: incompatible types
        Arrays.sort(array1, Comparator.comparingInt(tuple -> tuple[0]).thenComparingInt(tuple -> tuple[1])); 
        
        // Case (c): if downcast tuple[0] to ((int[])tuple)[0], then (b) works fine. 

更新：

1. 受到评论的启发，我很快意识到案例（b.1）实际上是无效的。(b.1) 中的 lambda 假设返回一个整数，而不是一个比较器。例如
   Arrays.sort(array1, (a1, a2) -> a1[0] == a2[0] ? 0 : 1);
2. 在所有其他场景中，我看到Comparator.<int[]>comparingInt(...)强制推理正确。

回答

Arrays.sort(array1, Comparator.<int[]>comparingInt(tuple -> tuple[0]).thenComparingInt(tuple -> tuple[1]));

Arrays.sort(array1,
            (a1, a2) -> a1[0] == a2[0] ?
                    Comparator.<int[]>comparingInt(tuple -> tuple[1]).compare(a1, a2) :
                    Comparator.<int[]>comparingInt(tuple -> tuple[0]).compare(a1, a2));