PyTorch中LSTM为何需初始化隐藏状态h0？其被覆盖仍需初始化？

Great question! Let's break this down clearly because there's a key misunderstanding here about how LSTM hidden states work vs. simple variable reassignment in code.

1. h0 isn't "overwritten" — it's the mandatory starting point of computation #

An LSTM calculates each time step's hidden state h_t using this core logic:
h_t = LSTM_cell(input_t, h_{t-1})

For the very first time step, there’s no h_{t-1} to use—so h0 fills that role as the initial hidden state. This isn’t like reassigning a variable (your a=0; a=4 example); it’s more like the first domino in a chain: without pushing that first domino (h0), none of the subsequent steps (h1, h2, etc.) can happen at all.

2. PyTorch auto-initializes h0, but manual initialization gives you critical control #

If you don’t explicitly pass h0 to PyTorch’s LSTM, the framework will automatically create a tensor of zeros for both h0 and c0 (the cell state). But manual initialization isn’t redundant—it lets you:

• Use learnable initial states: Instead of fixed zeros, you can make h0 a trainable parameter. The model can then learn the optimal starting state for your task, which helps with short sequences or specialized use cases where a zero start isn’t ideal.
• Carry over context between batches: For tasks like language modeling, you process text in batches. By passing the final hidden state of the previous batch as h0 for the next batch, you let the model maintain continuous context across longer texts—critical for capturing long-range dependencies.
• Set task-specific initial states: In some scenarios, you might want to seed the LSTM with a state aligned to your task (e.g., a positive/negative bias for sentiment analysis). Manual initialization lets you do this.

3. Skipping initialization entirely would break the model #

If PyTorch didn’t auto-initialize h0, your code would throw an error immediately. The LSTM cell requires an initial hidden state to start computing—there’s no way around it. The auto-zero initialization is just a sensible default, not proof that h0 is unnecessary.

To reframe your code analogy correctly, it’s not:

int a; a=0; a=4;

It’s more like:

h_init = 0  # Required starting point
h1 = lstm_step(input_1, h_init)
h2 = lstm_step(input_2, h1)

Here, h_init is a required input to generate h1—you can’t skip it and expect the computation to work.

内容的提问来源于stack exchange，提问作者sww