var _a = Object.prototype, toString = _a.toString, hasOwnProperty = _a.hasOwnProperty;
var previousComparisons = new Map();
/**
 * Performs a deep equality check on two JavaScript values, tolerating cycles.
 */
function equal(a, b) {
    try {
        return check(a, b);
    }
    finally {
        previousComparisons.clear();
    }
}
function check(a, b) {
    // If the two values are strictly equal, our job is easy.
    if (a === b) {
        return true;
    }
    // Object.prototype.toString returns a representation of the runtime type of
    // the given value that is considerably more precise than typeof.
    var aTag = toString.call(a);
    var bTag = toString.call(b);
    // If the runtime types of a and b are different, they could maybe be equal
    // under some interpretation of equality, but for simplicity and performance
    // we just return false instead.
    if (aTag !== bTag) {
        return false;
    }
    switch (aTag) {
        case '[object Array]':
            // Arrays are a lot like other objects, but we can cheaply compare their
            // lengths as a short-cut before comparing their elements.
            if (a.length !== b.length)
                return false;
        // Fall through to object case...
        case '[object Object]': {
            if (previouslyCompared(a, b))
                return true;
            var aKeys = Object.keys(a);
            var bKeys = Object.keys(b);
            // If `a` and `b` have a different number of enumerable keys, they
            // must be different.
            var keyCount = aKeys.length;
            if (keyCount !== bKeys.length)
                return false;
            // Now make sure they have the same keys.
            for (var k = 0; k < keyCount; ++k) {
                if (!hasOwnProperty.call(b, aKeys[k])) {
                    return false;
                }
            }
            // Finally, check deep equality of all child properties.
            for (var k = 0; k < keyCount; ++k) {
                var key = aKeys[k];
                if (!check(a[key], b[key])) {
                    return false;
                }
            }
            return true;
        }
        case '[object Error]':
            return a.name === b.name && a.message === b.message;
        case '[object Number]':
            // Handle NaN, which is !== itself.
            if (a !== a)
                return b !== b;
        // Fall through to shared +a === +b case...
        case '[object Boolean]':
        case '[object Date]':
            return +a === +b;
        case '[object RegExp]':
        case '[object String]':
            return a == "" + b;
        case '[object Map]':
        case '[object Set]': {
            if (a.size !== b.size)
                return false;
            if (previouslyCompared(a, b))
                return true;
            var aIterator = a.entries();
            var isMap = aTag === '[object Map]';
            while (true) {
                var info = aIterator.next();
                if (info.done)
                    break;
                // If a instanceof Set, aValue === aKey.
                var _a = info.value, aKey = _a[0], aValue = _a[1];
                // So this works the same way for both Set and Map.
                if (!b.has(aKey)) {
                    return false;
                }
                // However, we care about deep equality of values only when dealing
                // with Map structures.
                if (isMap && !check(aValue, b.get(aKey))) {
                    return false;
                }
            }
            return true;
        }
    }
    // Otherwise the values are not equal.
    return false;
}
function previouslyCompared(a, b) {
    // Though cyclic references can make an object graph appear infinite from the
    // perspective of a depth-first traversal, the graph still contains a finite
    // number of distinct object references. We use the previousComparisons cache
    // to avoid comparing the same pair of object references more than once, which
    // guarantees termination (even if we end up comparing every object in one
    // graph to every object in the other graph, which is extremely unlikely),
    // while still allowing weird isomorphic structures (like rings with different
    // lengths) a chance to pass the equality test.
    var bSet = previousComparisons.get(a);
    if (bSet) {
        // Return true here because we can be sure false will be returned somewhere
        // else if the objects are not equivalent.
        if (bSet.has(b))
            return true;
    }
    else {
        previousComparisons.set(a, bSet = new Set);
    }
    bSet.add(b);
    return false;
}

export default equal;
export { equal };
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