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events: simplify stack compare function #24744

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events: simplify stack compare function #24744

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events: simplify stack compare function
BridgeAR Nov 18, 2018
11a6984
fixup: improve comments
BridgeAR Nov 30, 2018
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41 changes: 20 additions & 21 deletions lib/events.js
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Original file line number Diff line number Diff line change
Expand Up @@ -103,30 +103,29 @@ EventEmitter.prototype.getMaxListeners = function getMaxListeners() {
return $getMaxListeners(this);
};

// Returns the longest sequence of `a` that fully appears in `b`,
// of length at least 3.
// This is a lazy approach but should work well enough, given that stack
// frames are usually unequal or otherwise appear in groups, and that
// we only run this code in case of an unhandled exception.
function longestSeqContainedIn(a, b) {
for (var len = a.length; len >= 3; --len) {
for (var i = 0; i < a.length - len; ++i) {
// Attempt to find a[i:i+len] in b
for (var j = 0; j < b.length - len; ++j) {
let matches = true;
for (var k = 0; k < len; ++k) {
if (a[i + k] !== b[j + k]) {
matches = false;
break;
}
// Returns the length and line number of the first sequence of `a` that fully
// appears in `b` with a length of at least 4.
function identicalSequenceRange(a, b) {
for (var i = 0; i < a.length - 3; i++) {
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I would cache a.length - 3 in a variable.

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why?

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I expect the value to be constant fold (but I did not check).

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a could be quite big, and it used to be slightly faster to cache the value if it's computed.

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a should normally be small (we currently only use this for stack frames) and this implementation should also be faster than the one before. If it's about performance, I could save a couple comparisons by using a simple for loop instead of indexOf (currently I check until the last entry but the last three entries are not interesting).

@bmeurer do values like these get constant fold?

// Find the first entry of b that matches the current entry of a.
const pos = b.indexOf(a[i]);
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if (pos !== -1) {
const rest = b.length - pos;
if (rest > 3) {
let len = 1;
const maxLen = Math.min(a.length - i, rest);
// Count the number of consecutive entries.
while (maxLen > len && a[i + len] === b[pos + len]) {
len++;
}
if (len > 3) {
return [len, i];
}
if (matches)
return [ len, i, j ];
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And for the same reason, returning j might not be used in this specific setup, but it’s part of having this be a more generic function.

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@BridgeAR BridgeAR Nov 30, 2018 •
edited
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pos translates to the former j but I recommend to change the signature when necessary and not to keep code in here that is currently unused.

}
}
}

return [ 0, 0, 0 ];
return [0, 0];
}

function enhanceStackTrace(err, own) {
Expand All @@ -135,9 +134,9 @@ function enhanceStackTrace(err, own) {
const errStack = err.stack.split('\n').slice(1);
const ownStack = own.stack.split('\n').slice(1);

const [ len, off ] = longestSeqContainedIn(ownStack, errStack);
const [ len, off ] = identicalSequenceRange(ownStack, errStack);
if (len > 0) {
ownStack.splice(off + 1, len - 1,
ownStack.splice(off + 1, len - 2,
' [... lines matching original stack trace ...]');
}
// Do this last, because it is the only operation with side effects.
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