Addition: dectests + implementation

scripts/dectest.jl

@@ -149,6 +149,7 @@ function translate(io, dectest_path)
         else
             test = _test(line)
             any(isspecial, test.operands) && continue
+            occursin("Unsupported", directives["rounding"]) && continue
             print_test(io, test, directives)
         end
     end

src/arithmetic.jl

@@ -7,22 +7,107 @@ Base.promote_rule(::Type{BigInt}, ::Type{Decimal}) = Decimal
 Base.:(+)(x::Decimal) = fix(x)
 Base.:(-)(x::Decimal) = fix(Decimal(!x.s, x.c, x.q))
 
-# Addition
-# To add, convert both decimals to the same exponent.
-# (If the exponents are different, use the smaller exponent
-# to make sure we're adding integers.)
 function Base.:(+)(x::Decimal, y::Decimal)
+    rmod = rounding(Decimal)
+
+    # Handle zero operands
+    if iszero(x) && iszero(y)
+        s = (x.s && y.s) || (rmod == RoundDown)
+        return Decimal(s, 0, min(x.q, y.q))
+    elseif iszero(x) # && !iszero(y)
+        return +(y)
+    elseif iszero(y) # && !iszero(x)
+        return +(x)
+    end
+
+    # Make sure that x.q ≥ y.q
     if x.q < y.q
         x, y = y, x
     end
-    # Here: x.q ≥ y.q
-    # a₁ * 10^q₁ + a₂ * 10^q₂ =
-    # (a₁ * 10^(q₁ - q₂) + a₂) * 10^q₂
-    #  ^^^^^^^^^^^^^^^^^ this is integer because q₁ ≥ q₂
-    q = x.q - y.q
-    c = (-1)^x.s * x.c * BigTen^q + (-1)^y.s * y.c
-    s = signbit(c)
-    return normalize(Decimal(s, abs(c), y.q))
+
+    # We are computing
+    #
+    #    (-1)^x.s * x.c * 10^x.q         + (-1)^y.s * y.c * 10^y.q =
+    #   [(-1)^x.s * x.c * 10^(x.q - y.q) + (-1)^y.s * y.c] * 10^y.q,
+    #
+    # where `10^(x.q - y.q)` is an integer because `x.q ≥ y.q`.
+    #
+    # Sometimes, `x.q` can be much larger than `y.q`, which would result in
+    # a huge coefficient `x.c * 10^(x.q - y.q)`. We want to prevent that.
+    #
+    # In the end, the resulting coefficient is still constraned by precision.
+    # Let `Ex, Ey` denote the adjusted exponents of `x` and `y`. If
+    #
+    #                       Ey < min(x.q - 1, Ex - prec - 1)
+    #   ndigits(y.c) + y.q - 1 < min(x.q - 1, ndigits(x.c) + x.q - 1 - prec - 1)
+    #       ndigits(y.c) + y.q < x.q + min(0, ndigits(x.c) - prec - 1)
+    #
+    # then we can replace `y` by a number `z` whose adjusted exponent `E`
+    # satisfies the equation
+    #
+    #   E = min(x.q - 1, Ex - prec - 1)
+    #
+    # and the result `x + z` equals `x + y` after rounding.
+    #
+    # From the above equation, we get that
+    #
+    #   ndigits(z.c) + z.q = x.q + min(0, ndigits(x.c) - precision - 1)
+    #
+    # Setting `z.c = 1`, we have `ndigits(z.c) = 1`, and thus
+    #
+    #   z.q = x.q - 1 + min(0, ndigits(x.c) - precision - 1).
+    #
+    # The problem of a huge coefficient is now gone because the difference
+    #
+    #   x.q - z.q = 1 - min(0, ndigits(x.c) - precision - 1)
+    #             = 1 + max(0, precision - ndigits(x.c) - 1)
+    #
+    # is upper-bounded by precision.
+
+    v = x.q + min(0, ndigits(x.c) - precision(Decimal) - 1)
+    if ndigits(y.c) + y.q < v
+        y = Decimal(y.s, BigOne, v - 1)
+    end
+
+    xc = x.c * BigTen ^ (x.q - y.q)
+    yc = y.c
+    q = y.q
+
+    # The addition now amounts to
+    #
+    #   ((-1)^x.s * xc + (-1)^y.s * yc) * 10^q
+    #    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
+    #
+    # We take some extra steps to compute the addition highlighted above while
+    # allocating as few BigInts as possible
+
+    # If the signs are equal, we compute either `x + y` or `-x - y`, which can
+    # be generally written as `s * (x + y)`, where `s` is the sign of `x` and `y`
+    if x.s == y.s
+        return fix(Decimal(x.s, xc + yc, q))
+    end
+
+    # If the signs `x.s` and `y.s` differ, we compute either `xc - yc` or
+    # `yc - xc`. There are four possibilities:
+    #
+    #   Signs
+    #   x   y   Magnit.   Result
+    #   -----------------------------------
+    #   +   -   xc > yc   +(xc - yc) * 10^q
+    #   +   -   xc ≤ yc   -(yc - xc) * 10^q
+    #   -   +   xc > yc   -(xc - yc) * 10^q
+    #   -   +   xc ≤ yc   +(yc - xc) * 10^q
+    #
+    # If the result is zero (i.e., `xc == yc`), it should be negative if the
+    # rounding mode is `RoundDown` and positive otherwise.
+
+    if xc == yc
+        return fix(Decimal(rmod === RoundDown, BigZero, q))
+    elseif xc > yc
+        return fix(Decimal(x.s, xc - yc, q))
+    else
+        return fix(Decimal(y.s, yc - xc, q))
+    end
 end
 
 # Subtraction

src/bigint.jl

@@ -57,3 +57,17 @@
 
     return x, q
 end
+
+if VERSION < v"1.9.0"
+    # Taken from PR #41246 at JuliaLang/julia
+    # The methods are present from 1.9 so if we bump the required Julia version
+    # to 1.9, we can remove this block
+
+    function Base.div(x::Integer, y::Integer, ::typeof(RoundFromZero))
+        signbit(x) == signbit(y) ? div(x, y, RoundUp) : div(x, y, RoundDown)
+    end
+
+    function Base.rem(x, y, ::typeof(RoundFromZero))
+        signbit(x) == signbit(y) ? rem(x, y, RoundUp) : rem(x, y, RoundDown)
+    end
+end