C++ JSString::setNonInlineChars方法代码示例

JSFlatString *
JSRope::flattenInternal(ExclusiveContext *maybecx)
{
    /*
     * Perform a depth-first dag traversal, splatting each node's characters
     * into a contiguous buffer. Visit each rope node three times:
     *   1. record position in the buffer and recurse into left child;
     *   2. recurse into the right child;
     *   3. transform the node into a dependent string.
     * To avoid maintaining a stack, tree nodes are mutated to indicate how many
     * times they have been visited. Since ropes can be dags, a node may be
     * encountered multiple times during traversal. However, step 3 above leaves
     * a valid dependent string, so everything works out.
     *
     * While ropes avoid all sorts of quadratic cases with string
     * concatenation, they can't help when ropes are immediately flattened.
     * One idiomatic case that we'd like to keep linear (and has traditionally
     * been linear in SM and other JS engines) is:
     *
     *   while (...) {
     *     s += ...
     *     s.flatten
     *   }
     *
     * To do this, when the buffer for a to-be-flattened rope is allocated, the
     * allocation size is rounded up. Then, if the resulting flat string is the
     * left-hand side of a new rope that gets flattened and there is enough
     * capacity, the rope is flattened into the same buffer, thereby avoiding
     * copying the left-hand side. Clearing the 'extensible' bit turns off this
     * optimization. This is necessary, e.g., when the JSAPI hands out the raw
     * null-terminated char array of a flat string.
     *
     * N.B. This optimization can create chains of dependent strings.
     */
    const size_t wholeLength = length();
    size_t wholeCapacity;
    CharT *wholeChars;
    JSString *str = this;
    CharT *pos;

    /*
     * JSString::flattenData is a tagged pointer to the parent node.
     * The tag indicates what to do when we return to the parent.
     */
    static const uintptr_t Tag_Mask = 0x3;
    static const uintptr_t Tag_FinishNode = 0x0;
    static const uintptr_t Tag_VisitRightChild = 0x1;

    AutoCheckCannotGC nogc;

    /* Find the left most string, containing the first string. */
    JSRope *leftMostRope = this;
    while (leftMostRope->leftChild()->isRope())
        leftMostRope = &leftMostRope->leftChild()->asRope();

    if (leftMostRope->leftChild()->isExtensible()) {
        JSExtensibleString &left = leftMostRope->leftChild()->asExtensible();
        size_t capacity = left.capacity();
        if (capacity >= wholeLength && left.hasTwoByteChars() == IsSame<CharT, jschar>::value) {
            /*
             * Simulate a left-most traversal from the root to leftMost->leftChild()
             * via first_visit_node
             */
            JS_ASSERT(str->isRope());
            while (str != leftMostRope) {
                if (b == WithIncrementalBarrier) {
                    JSString::writeBarrierPre(str->d.s.u2.left);
                    JSString::writeBarrierPre(str->d.s.u3.right);
                }
                JSString *child = str->d.s.u2.left;
                JS_ASSERT(child->isRope());
                str->setNonInlineChars(left.nonInlineChars<CharT>(nogc));
                child->d.u1.flattenData = uintptr_t(str) | Tag_VisitRightChild;
                str = child;
            }
            if (b == WithIncrementalBarrier) {
                JSString::writeBarrierPre(str->d.s.u2.left);
                JSString::writeBarrierPre(str->d.s.u3.right);
            }
            str->setNonInlineChars(left.nonInlineChars<CharT>(nogc));
            wholeCapacity = capacity;
            wholeChars = const_cast<CharT *>(left.nonInlineChars<CharT>(nogc));
            pos = wholeChars + left.d.u1.length;
            JS_STATIC_ASSERT(!(EXTENSIBLE_FLAGS & DEPENDENT_FLAGS));
            left.d.u1.flags ^= (EXTENSIBLE_FLAGS | DEPENDENT_FLAGS);
            left.d.s.u3.base = (JSLinearString *)this;  /* will be true on exit */
            StringWriteBarrierPostRemove(maybecx, &left.d.s.u2.left);
            StringWriteBarrierPost(maybecx, (JSString **)&left.d.s.u3.base);
            goto visit_right_child;
        }
    }

    if (!AllocChars(maybecx, wholeLength, &wholeChars, &wholeCapacity))
        return nullptr;

    pos = wholeChars;
    first_visit_node: {
        if (b == WithIncrementalBarrier) {
            JSString::writeBarrierPre(str->d.s.u2.left);
            JSString::writeBarrierPre(str->d.s.u3.right);
//.........这里部分代码省略.........

JSFlatString*
JSRope::flattenInternal(ExclusiveContext* maybecx)
{
    /*
     * Consider the DAG of JSRopes rooted at this JSRope, with non-JSRopes as
     * its leaves. Mutate the root JSRope into a JSExtensibleString containing
     * the full flattened text that the root represents, and mutate all other
     * JSRopes in the interior of the DAG into JSDependentStrings that refer to
     * this new JSExtensibleString.
     *
     * If the leftmost leaf of our DAG is a JSExtensibleString, consider
     * stealing its buffer for use in our new root, and transforming it into a
     * JSDependentString too. Do not mutate any of the other leaves.
     *
     * Perform a depth-first dag traversal, splatting each node's characters
     * into a contiguous buffer. Visit each rope node three times:
     *   1. record position in the buffer and recurse into left child;
     *   2. recurse into the right child;
     *   3. transform the node into a dependent string.
     * To avoid maintaining a stack, tree nodes are mutated to indicate how many
     * times they have been visited. Since ropes can be dags, a node may be
     * encountered multiple times during traversal. However, step 3 above leaves
     * a valid dependent string, so everything works out.
     *
     * While ropes avoid all sorts of quadratic cases with string concatenation,
     * they can't help when ropes are immediately flattened. One idiomatic case
     * that we'd like to keep linear (and has traditionally been linear in SM
     * and other JS engines) is:
     *
     *   while (...) {
     *     s += ...
     *     s.flatten
     *   }
     *
     * Two behaviors accomplish this:
     *
     * - When the leftmost non-rope in the DAG we're flattening is a
     *   JSExtensibleString with sufficient capacity to hold the entire
     *   flattened string, we just flatten the DAG into its buffer. Then, when
     *   we transform the root of the DAG from a JSRope into a
     *   JSExtensibleString, we steal that buffer, and change the victim from a
     *   JSExtensibleString to a JSDependentString. In this case, the left-hand
     *   side of the string never needs to be copied.
     *
     * - Otherwise, we round up the total flattened size and create a fresh
     *   JSExtensibleString with that much capacity. If this in turn becomes the
     *   leftmost leaf of a subsequent flatten, we will hopefully be able to
     *   fill it, as in the case above.
     *
     * Note that, even though the code for creating JSDependentStrings avoids
     * creating dependents of dependents, we can create that situation here: the
     * JSExtensibleStrings we transform into JSDependentStrings might have
     * JSDependentStrings pointing to them already. Stealing the buffer doesn't
     * change its address, only its owning JSExtensibleString, so all chars()
     * pointers in the JSDependentStrings are still valid.
     */
    const size_t wholeLength = length();
    size_t wholeCapacity;
    CharT* wholeChars;
    JSString* str = this;
    CharT* pos;

    /*
     * JSString::flattenData is a tagged pointer to the parent node.
     * The tag indicates what to do when we return to the parent.
     */
    static const uintptr_t Tag_Mask = 0x3;
    static const uintptr_t Tag_FinishNode = 0x0;
    static const uintptr_t Tag_VisitRightChild = 0x1;

    AutoCheckCannotGC nogc;

    /* Find the left most string, containing the first string. */
    JSRope* leftMostRope = this;
    while (leftMostRope->leftChild()->isRope())
        leftMostRope = &leftMostRope->leftChild()->asRope();

    if (leftMostRope->leftChild()->isExtensible()) {
        JSExtensibleString& left = leftMostRope->leftChild()->asExtensible();
        size_t capacity = left.capacity();
        if (capacity >= wholeLength && left.hasTwoByteChars() == IsSame<CharT, char16_t>::value) {
            /*
             * Simulate a left-most traversal from the root to leftMost->leftChild()
             * via first_visit_node
             */
            MOZ_ASSERT(str->isRope());
            while (str != leftMostRope) {
                if (b == WithIncrementalBarrier) {
                    JSString::writeBarrierPre(str->d.s.u2.left);
                    JSString::writeBarrierPre(str->d.s.u3.right);
                }
                JSString* child = str->d.s.u2.left;
                MOZ_ASSERT(child->isRope());
                str->setNonInlineChars(left.nonInlineChars<CharT>(nogc));
                child->d.u1.flattenData = uintptr_t(str) | Tag_VisitRightChild;
                str = child;
            }
            if (b == WithIncrementalBarrier) {
                JSString::writeBarrierPre(str->d.s.u2.left);
                JSString::writeBarrierPre(str->d.s.u3.right);
//.........这里部分代码省略.........