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Fix doubled keystrokes after session exits in interactive list
The EXIT packet handler in attach() didn't set exitHandled, so the
socket close event called onExit a second time, which added a
duplicate stdin listener via resumeTui → setupInput.
Added regression test that types after session exit and verifies
characters aren't doubled.
Nathan Herald
(Mar 26, 2026, 2:31 PM +0100)
9f15b9d5
ef1c8a4a
+48
-1
+1
-1
package.json
+1
-1
package.json
+1
src/client.ts
+1
src/client.ts
+46
tests/tui.test.ts
+46
tests/tui.test.ts
···
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);
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it(
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"after session exits and returning to list, keystrokes are not doubled",
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async () => {
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const sessionDir = makeSessionDir();
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const name = uniqueName();
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// A session that exits immediately after printing a marker
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const { pid } = await createBackgroundSession(
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sessionDir,
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name,
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"sh",
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["-c", "echo WILL_EXIT; exec cat"],
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os.tmpdir()
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);
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bgPids.push(pid);
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const tui = createTuiSession(sessionDir);
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// Attach
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await tui.waitForText(name, 10000);
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tui.press("return");
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await tui.waitForText("WILL_EXIT", 10000);
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// Make the session exit (Ctrl+D sends EOF to cat)
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tui.sendKeys("\x04");
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// Wait for return to list
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await tui.waitForText("Create new session...", 10000);
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// Type a filter string and verify each character appears exactly once.
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// If stdin has duplicate listeners, each keystroke fires twice and
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// the filter will contain doubled characters (e.g. "xxyz" instead of "xyz").
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tui.type("x");
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await new Promise(r => setTimeout(r, 300));
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tui.type("y");
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await new Promise(r => setTimeout(r, 300));
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tui.type("z");
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await new Promise(r => setTimeout(r, 300));
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const ss = tui.screenshot();
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expect(ss.text).toContain("xyz");
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expect(ss.text).not.toContain("xxyyzz");
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},
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20000
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);
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it(
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"exited session shows as exited when returning to list during cleanup window",
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async () => {
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const sessionDir = makeSessionDir();