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When a LangGraph run resumes after interrupt(), the node containing that call starts again from its first statement. The call then returns the value you supplied through Command(resume=...), and the node continues. That restart is expected behavior—not, by itself, evidence of an accidental graph loop.
Why does the interrupted node start over?
interrupt() pauses graph execution and exposes a payload for the caller to handle. LangGraph persists the paused graph state through a checkpointer. When you resume, the runtime re-enters the node from its beginning; it does not continue at the exact Python instruction after interrupt().
On the resumed attempt, the node executes its earlier statements again. When it reaches the same interrupt() call, that call returns the value supplied in Command(resume=...) rather than pausing again. Statements after the call then run with that value. LangChain’s official interrupt guide describes this restart behavior explicitly.
What should a resume call look like?
The graph needs a checkpointer to persist the paused state, and the resume invocation must use the same configured thread_id as the initial invocation. That lets LangGraph locate and continue the paused thread. Using a different ID starts a separate thread instead.
from langgraph.types import Command, interrupt
def approval_node(state):
# Runs on the initial attempt and again after resume.
request = build_approval_request(state)
approved = interrupt(request)
# Runs after the resume value is returned.
return {"approved": approved}
# Initial invocation pauses at interrupt().
result = graph.invoke(
input_data,
config={"configurable": {"thread_id": "case-123"}},
)
# Resume the same thread.
result = graph.invoke(
Command(resume=True),
config={"configurable": {"thread_id": "case-123"}},
)Here, True becomes the return value of interrupt(request) on resumption. The example assumes graph was compiled with a checkpointer.
How can you prevent duplicate side effects?
The restart applies to the node containing the interrupt; it does not mean every node in the workflow necessarily runs again. The main risk is a side effect in that node before the interrupt, because that code executes again on resume.
- Keep pre-interrupt work free of external effects. Build a request or calculate values before the pause, but avoid sending messages, writing records, or making irreversible external changes there where possible.
- Make unavoidable work idempotent. For example, an application-level idempotency key can let an external operation recognize a retry. That is an application design technique, not a LangGraph-specific guarantee.
- Move the effect after the interrupt. Then it runs only after the resume value is returned and the node proceeds.
- Put the effect in a separate node. This makes the execution boundary explicit and can simplify reasoning about retries.
These patterns reduce the chance that a resumed node repeats an externally visible action. Choose based on whether the operation can safely be repeated and when it needs to happen relative to human input.
What can go wrong with multiple interrupts or exception handling?
Keep multiple interrupt calls in the same order
If a node contains more than one interrupt() call, preserve their order between the original attempt and the resumed attempt. LangGraph matches resume values by position, so changing the order can associate a value with the wrong interrupt. The Python API reference documents the interrupt API.
Do not swallow the pause signal
An interrupt uses a special control-flow exception that LangGraph handles to pause execution. A broad try/except surrounding interrupt() can catch and swallow that signal, interfering with the pause. Keep handling for ordinary application errors separate from the interrupt call.
How to tell a restart from an accidental loop
- If the node runs again after a resume and proceeds with the supplied resume value, that matches the documented interrupt behavior.
- If code before
interrupt()repeats, check whether that work is safe to run again; it is part of the restarted node. - If the run does not continue the paused thread, verify that the graph uses a checkpointer and that the resume call uses the original
thread_id. - If multiple interrupts behave unexpectedly, confirm their order is unchanged and that exception handling is not catching the pause signal.
LangGraph’s interrupt behavior can be version-sensitive. The official documentation and API reference are the best places to confirm details for the version installed in your application.
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