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Unary minus of a constant does make the number signed. 

It is wrong to assume that the result of a unary minus is signed.
Doing so can cause incorrect results in subsequent math.
This commit is contained in:
Stephen Williams 2008-05-09 14:18:39 -07:00
parent 8e1f22cd82
commit a1495bb007
1 changed files with 3 additions and 4 deletions
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7
elab_expr.cc
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@ -1866,16 +1866,15 @@ NetExpr* PEUnary::elaborate_expr(Design*des, NetScope*scope,
if (expr_wid > 0) if (expr_wid > 0)
val = pad_to_width(val, expr_wid); val = pad_to_width(val, expr_wid);
/* When taking the - of a number, turn it into a /* When taking the - of a number, extend it one
signed expression and extend it one bit to bit to accommodate a possible sign bit. */
accommodate a possible sign bit. */
verinum zero (verinum::V0, val.len()+1, val.has_len()); verinum zero (verinum::V0, val.len()+1, val.has_len());
zero.has_sign(val.has_sign()); zero.has_sign(val.has_sign());
verinum nval = zero - val; verinum nval = zero - val;
if (val.has_len()) if (val.has_len())
nval = verinum(nval, val.len()); nval = verinum(nval, val.len());
nval.has_sign(true); nval.has_sign(val.has_sign());
tmp = new NetEConst(nval); tmp = new NetEConst(nval);
tmp->set_line(*this); tmp->set_line(*this);
delete ip; delete ip;