AstNode* replaceCaseFastRecurse(AstNode* cexprp, int msb, uint32_t upperValue) {
if (msb<0) {
// There's no space for a IF. We know upperValue is thus down to a specific
// exact value, so just return the tree value
// Note can't clone here, as we're going to check for equivelence above
return m_valueItem[upperValue];
}
else {
// Make left and right subtrees
// cexpr[msb:lsb] == 1
AstNode* tree0p = replaceCaseFastRecurse(cexprp, msb-1, upperValue | 0);
AstNode* tree1p = replaceCaseFastRecurse(cexprp, msb-1, upperValue | (1UL<<msb));
if (tree0p == tree1p) {
// Same logic on both sides, so we can just return one of 'em
return tree0p;
}
// We could have a "checkerboard" with A B A B, we can use the same IF on both edges
bool same = true;
for (uint32_t a=upperValue,
b=(upperValue|(1UL<<msb));
a < (upperValue|(1UL<<msb));
a++, b++) {
if (m_valueItem[a] != m_valueItem[b]) { same=false; break; }
}
if (same) {
tree1p->deleteTree(); tree1p=NULL;
return tree0p;
}
// Must have differing logic, so make a selection
// Case expressions can't be linked twice, so clone them
if (tree0p && !tree0p->user3()) tree0p = tree0p->cloneTree(true);
if (tree1p && !tree1p->user3()) tree1p = tree1p->cloneTree(true);
// Alternate scheme if we ever do multiple bits at a time:
//V3Number nummask (cexprp->fileline(), cexprp->width(), (1UL<<msb));
//AstNode* and1p = new AstAnd(cexprp->fileline(), cexprp->cloneTree(false),
// new AstConst(cexprp->fileline(), nummask));
AstNode* and1p = new AstSel(cexprp->fileline(), cexprp->cloneTree(false),
msb, 1);
AstNode* eqp = new AstNeq(cexprp->fileline(),
new AstConst(cexprp->fileline(), 0),
and1p);
AstIf* ifp = new AstIf(cexprp->fileline(), eqp, tree1p, tree0p);
ifp->user3(1); // So we don't bother to clone it
return ifp;
}
}
