ArrayList的add方法工作原理及TableRow构造调用作用问询

Alright, let’s unpack this piece by piece to clear up your confusion:

First, let’s break down the components:

• Constructor call note: Your TableRow class has a constructor that requires three parameters: String symbol, int address, int index. The code here only passes two arguments, which will trigger a compile error—this is almost certainly a typo. I suspect you intended to pass three values, like new TableRow(parts[1], Integer.parseInt(parts[2]), Integer.parseInt(parts[0])) to match the constructor’s signature.
• Parameter breakdown:
  • parts[1]: This string value gets assigned to the symbol field of the new TableRow object (likely representing a symbol name in your symbol table).
  • Integer.parseInt(parts[2], Integer.parseInt(parts[0])): This is a nested integer conversion. The inner Integer.parseInt(parts[0]) turns the string parts[0] into an integer, which acts as the number base (radix) for the outer conversion. The outer call then converts parts[2] from that base to a decimal integer—this would typically map to the address field in your constructor.
• Core function: This line creates a brand-new instance of the TableRow class, initializing its symbol, address, and index fields with values pulled from the parts array.
• Return value: The entire expression returns a reference (memory address) to the newly created TableRow object. This reference is what gets passed directly to the ArrayList.add() method.

• Internal mechanism: ArrayList uses a resizable array under the hood. When you call add():
  1. It first checks if the underlying array has enough empty space for the new element.
  2. If space is available, it places the TableRow object reference into the next open position in the array.
  3. If the array is full, it automatically resizes (usually doubling its capacity), copies existing elements to the new larger array, then adds the new element.
• Return value: The add() method returns a boolean value. For ArrayList, this will always return true (since it’s an unbounded collection that allows duplicates—only a catastrophic out-of-memory error would cause a failure, which is an extreme edge case).

内容的提问来源于stack exchange，提问作者Mercury1337