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fix(mcp): clear stale thread interrupt before MCP discovery

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Fixes #9930

When an agent session is interrupted (Ctrl+C or gateway timeout), the
current thread's interrupt flag is set in _interrupted_threads. asyncio
executor threads are pooled and reused across sessions, so a thread that
carried an interrupt flag from a prior session will immediately cancel
any new asyncio work dispatched to it — including MCP server discovery.

Fix: in register_mcp_servers(), temporarily clear the interrupt flag on
the current thread before running _discover_all(), then restore it
afterward in a finally block so the original interrupt state is not lost.
This commit is contained in:
AJV20 2026-04-15 08:01:15 -04:00 • committed by Teknium
parent b7a97cd44f
commit 9575bce6ca
1 changed files with 13 additions and 1 deletions
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14
tools/mcp_tool.py
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@ -2922,7 +2922,19 @@ def register_mcp_servers(servers: Dict[str, dict]) -> List[str]:
# Per-server timeouts are handled inside _discover_and_register_server.
# The outer timeout is generous: 120s total for parallel discovery.
_run_on_mcp_loop(_discover_all(), timeout=120)
#
# Temporarily clear the interrupt flag on the current thread so that MCP
# discovery is never cancelled by a stale interrupt from a prior agent
# session (executor threads get reused and may carry old interrupt state).
from tools.interrupt import is_interrupted as _is_interrupted, set_interrupt as _set_interrupt
_was_interrupted = _is_interrupted()
if _was_interrupted:
_set_interrupt(False)
try:
_run_on_mcp_loop(_discover_all(), timeout=120)
finally:
if _was_interrupted:
_set_interrupt(True)
# Log a summary so ACP callers get visibility into what was registered.
with _lock: