> 've started developing SymbolicNim
That is an interesting project. I tried some symbolic math myself when I was a
young boy but fully failed and never tried again. : -)
I had just a short look at your code. Four spaces for indentation -- live would
be so easy with tabs...
Some of your code looks a bit funny to me, but well it is hot hear and mybe I
am a bit stupid today...
proc isHalfMultiple(r: SymNumberType): bool =
if r.den == 2 or r.den == 1: return true
return false
proc isHalfMultiple(r: SymNumberType): bool =
r.den == 2 or r.den == 1
Run
proc pow*(r: SymNumberType, e: int): SymNumberType =
if e < 0:
result.num = r.den ^ (-e)
result.den = r.num ^ (-e)
else:
result.num = r.num ^ e
result.den = r.den ^ e
proc pow*(r: SymNumberType, e: int): SymNumberType =
result.num = r.den ^ (-e.abs)
result.den = r.num ^ (-e.abs)
Run
proc contains*(tree: SymbolicExpression, kind: ExprKind): bool
{.noSideEffect.} =
if tree.children.len == 0:
return false
else:
for child in tree.children:
if child.kind == kind:
return true
return false
proc contains*(tree: SymbolicExpression, kind: ExprKind): bool
{.noSideEffect.} =
for child in tree.children:
if child.kind == kind:
return true
Run
proc survey(trees: openArray[SymbolicExpression]): SurveyObject =
for e in ExprKind:
result[e] = (count: 0, indexes: newSeqOfCap[int]((trees.len /
2).toInt))
for i in 0 .. trees.high:
case trees[i].kind
of exprConstant:
inc result[exprConstant].count
result[exprConstant].indexes.add(i)
of exprVariable:
inc result[exprVariable].count
result[exprVariable].indexes.add(i)
of exprFactors:
inc result[exprFactors].count
result[exprFactors].indexes.add(i)
of exprTerms:
inc result[exprTerms].count
result[exprTerms].indexes.add(i)
of exprExponent:
inc result[exprExponent].count
result[exprExponent].indexes.add(i)
of exprFuncCall:
inc result[exprFuncCall].count
result[exprFuncCall].indexes.add(i)
proc survey(trees: openArray[SymbolicExpression]): SurveyObject =
for e in ExprKind:
result[e] = (count: 0, indexes: newSeqOfCap[int]((trees.len /
2).toInt))
for i in 0 .. trees.high:
let x = trees[i].kind
inc result[x].count
result[x].indexes.add(i)
Run
What do you think?
