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; } }