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------------------------------------------------------------------------------
-- --
-- GNAT RUN-TIME COMPONENTS --
-- --
-- S Y S T E M . I M A G E _ F --
-- --
-- B o d y --
-- --
-- Copyright (C) 2020-2022, Free Software Foundation, Inc. --
-- --
-- GNAT is free software; you can redistribute it and/or modify it under --
-- terms of the GNU General Public License as published by the Free Soft- --
-- ware Foundation; either version 3, or (at your option) any later ver- --
-- sion. GNAT is distributed in the hope that it will be useful, but WITH- --
-- OUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY --
-- or FITNESS FOR A PARTICULAR PURPOSE. --
-- --
-- As a special exception under Section 7 of GPL version 3, you are granted --
-- additional permissions described in the GCC Runtime Library Exception, --
-- version 3.1, as published by the Free Software Foundation. --
-- --
-- You should have received a copy of the GNU General Public License and --
-- a copy of the GCC Runtime Library Exception along with this program; --
-- see the files COPYING3 and COPYING.RUNTIME respectively. If not, see --
-- <http://www.gnu.org/licenses/>. --
-- --
-- GNAT was originally developed by the GNAT team at New York University. --
-- Extensive contributions were provided by Ada Core Technologies Inc. --
-- --
------------------------------------------------------------------------------
with System.Image_I;
with System.Img_Util; use System.Img_Util;
package body System.Image_F is
Maxdigs : constant Natural := Int'Width - 2;
-- Maximum number of decimal digits that can be represented in an Int.
-- The "-2" accounts for the sign and one extra digit, since we need the
-- maximum number of 9's that can be represented, e.g. for the 64-bit case,
-- Integer_64'Width is 20 since the maximum value is approximately 9.2E+18
-- and has 19 digits, but the maximum number of 9's that can be represented
-- in Integer_64 is only 18.
-- The first prerequisite of the implementation is that the first scaled
-- divide does not overflow, which means that the absolute value of the
-- input X must always be smaller than 10**Maxdigs * 2**(Int'Size - 1).
-- Otherwise Constraint_Error is raised by the scaled divide operation.
-- The second prerequisite of the implementation is that the computation
-- of the operands does not overflow when the small is neither an integer
-- nor the reciprocal of an integer, which means that its numerator and
-- denominator must be smaller than 10**(2*Maxdigs-1) / 2**(Int'Size - 1)
-- if the small is larger than 1, and smaller than 2**(Int'Size - 1) / 10
-- if the small is smaller than 1.
package Image_I is new System.Image_I (Int);
procedure Set_Image_Integer
(V : Int;
S : in out String;
P : in out Natural)
renames Image_I.Set_Image_Integer;
-- The following section describes a specific implementation choice for
-- performing base conversions needed for output of values of a fixed
-- point type T with small T'Small. The goal is to be able to output
-- all values of fixed point types with a precision of 64 bits and a
-- small in the range 2.0**(-63) .. 2.0**63. The reasoning can easily
-- be adapted to fixed point types with a precision of 32 or 128 bits.
-- The chosen algorithm uses fixed precision integer arithmetic for
-- reasons of simplicity and efficiency. It is important to understand
-- in what ways the most simple and accurate approach to fixed point I/O
-- is limiting, before considering more complicated schemes.
-- Without loss of generality assume T has a range (-2.0**63) * T'Small
-- .. (2.0**63 - 1) * T'Small, and is output with Aft digits after the
-- decimal point and T'Fore - 1 before. If T'Small is integer, or
-- 1.0 / T'Small is integer, let S = T'Small.
-- The idea is to convert a value X * S of type T to a 64-bit integer value
-- Q equal to 10.0**D * (X * S) rounded to the nearest integer, using only
-- a scaled integer divide of the form
-- Q = (X * Y) / Z,
-- where the variables X, Y, Z are 64-bit integers, and both multiplication
-- and division are done using full intermediate precision. Then the final
-- decimal value to be output is
-- Q * 10**(-D)
-- This value can be written to the output file or to the result string
-- according to the format described in RM A.3.10. The details of this
-- operation are omitted here.
-- A 64-bit value can represent all integers with 18 decimal digits, but
-- not all with 19 decimal digits. If the total number of requested ouput
-- digits (Fore - 1) + Aft is greater than 18 then, for purposes of the
-- conversion, Aft is adjusted to 18 - (Fore - 1). In that case, trailing
-- zeros can complete the output after writing the first 18 significant
-- digits, or the technique described in the next section can be used.
-- In addition, D cannot be smaller than -18, in order for 10.0**(-D) to
-- fit in a 64-bit integer.
-- The final expression for D is
-- D = Integer'Max (-18, Integer'Min (Aft, 18 - (Fore - 1)));
-- For Y and Z the following expressions can be derived:
-- Q = X * S * (10.0**D) = (X * Y) / Z
-- If S is an integer greater than or equal to one, then Fore must be at
-- least 20 in order to print T'First, which is at most -2.0**63. This
-- means that D < 0, so use
-- (1) Y = -S and Z = -10**(-D)
-- If 1.0 / S is an integer greater than one, use
-- (2) Y = -10**D and Z = -(1.0 / S), for D >= 0
-- or
-- (3) Y = -1 and Z = -(1.0 / S) * 10**(-D), for D < 0
-- Negative values are used for nominator Y and denominator Z, so that S
-- can have a maximum value of 2.0**63 and a minimum of 2.0**(-63). For
-- -(1.0 / S) in -1 .. -9, Fore will still be 20, and D will be negative,
-- as (-2.0**63) / -9 is greater than 10**18. In these cases there is room
-- in the denominator for the extra decimal scaling required, so case (3)
-- will not overflow.
-- In fact this reasoning can be generalized to most S which are the ratio
-- of two integers with bounded magnitude. Let S = Num / Den and rewrite
-- case (1) above where Den = 1 into
-- (1b) Y = -Num and Z = -Den * 10**(-D)
-- Suppose that Num corresponds to the maximum value of -D, i.e. 18 and
-- 10**(-D) = 10**18. If you change Den into 10, then S becomes 10 times
-- smaller and, therefore, Fore is decremented by 1, which means that -D
-- is as well, provided that Fore was initially not larger than 37, so the
-- multiplication for Z still does not overflow. But you need to reach 10
-- to trigger this effect, which means that a leeway of 10 is required, so
-- let's restrict this to a Num for which 10**(-D) <= 10**17. To summarize
-- this case, if S is the ratio of two integers with
-- Den < Num <= B1
-- where B1 is a fixed limit, then case (1b) does not overflow. B1 can be
-- taken as the largest integer Small such that D = -17, i.e. Fore = 36,
-- which means that B1 * 2.0**63 must be smaller than 10**35.
-- Let's continue and rewrite case (2) above when Num = 1 into
-- (2b) Y = -Num * 10**D and Z = -Den, for D >= 0
-- Note that D <= 18 - (Fore - 1) and Fore >= 2 so D <= 17, thus you can
-- safely change Num into 10 in the product, but then S becomes 10 times
-- larger and, therefore, Fore is incremented by 1, which means that D is
-- decremented by 1 so you again have a product lesser or equal to 10**17.
-- To sum up, if S is the ratio of two integers with
-- Num <= Den * S0
-- where S0 is the largest Small such that D >= 0, then case (2b) does not
-- overflow.
-- Let's conclude and rewrite case (3) above when Num = 1 into
-- (3b) Y = -Num and Z = -Den * 10**(-D), for D < 0
-- As explained above, this occurs only if both S0 < S < 1 and D = -1 and
-- is preserved if you scale up Num and Den simultaneously, what you can
-- do until Den * 10 tops the upper bound. To sum up, if S is the ratio of
-- two integers with
-- Den * S0 < Num < Den <= B2
-- where B2 is a fixed limit, then case (3b) does not overflow. B2 can be
-- taken as the largest integer such that B2 * 10 is smaller than 2.0**63.
-- The conclusion is that the algorithm works if the small is the ratio of
-- two integers in the range 1 .. 2**63 if either is equal to 1, or of two
-- integers in the range 1 .. B1 if the small is larger than 1, or of two
-- integers in the range 1 .. B2 if the small is smaller than 1.
-- Using a scaled divide which truncates and returns a remainder R,
-- another K trailing digits can be calculated by computing the value
-- (R * (10.0**K)) / Z using another scaled divide. This procedure
-- can be repeated to compute an arbitrary number of digits in linear
-- time and storage. The last scaled divide should be rounded, with
-- a possible carry propagating to the more significant digits, to
-- ensure correct rounding of the unit in the last place.
-----------------
-- Image_Fixed --
-----------------
procedure Image_Fixed
(V : Int;
S : in out String;
P : out Natural;
Num : Int;
Den : Int;
For0 : Natural;
Aft0 : Natural)
is
pragma Assert (S'First = 1);
begin
-- Add space at start for non-negative numbers
if V >= 0 then
S (1) := ' ';
P := 1;
else
P := 0;
end if;
Set_Image_Fixed (V, S, P, Num, Den, For0, Aft0, 1, Aft0, 0);
end Image_Fixed;
---------------------
-- Set_Image_Fixed --
---------------------
procedure Set_Image_Fixed
(V : Int;
S : in out String;
P : in out Natural;
Num : Int;
Den : Int;
For0 : Natural;
Aft0 : Natural;
Fore : Natural;
Aft : Natural;
Exp : Natural)
is
pragma Assert (Num < 0 and then Den < 0);
-- Accept only negative numbers to allow -2**(Int'Size - 1)
A : constant Natural :=
Boolean'Pos (Exp > 0) * Aft0 + Natural'Max (Aft, 1) + 1;
-- Number of digits after the decimal point to be computed. If Exp is
-- positive, we need to compute Aft decimal digits after the first non
-- zero digit and we are guaranteed there is at least one in the first
-- Aft0 digits (unless V is zero). In both cases, we compute one more
-- digit than requested so that Set_Decimal_Digits can round at Aft.
D : constant Integer :=
Integer'Max (-Maxdigs, Integer'Min (A, Maxdigs - (For0 - 1)));
Y : constant Int := Num * 10**Integer'Max (0, D);
Z : constant Int := Den * 10**Integer'Max (0, -D);
-- See the description of the algorithm above
AF : constant Natural := A - D;
-- Number of remaining digits to be computed after the first round. It
-- is larger than A if the first round does not compute all the digits
-- before the decimal point, i.e. (For0 - 1) larger than Maxdigs.
N : constant Natural := 1 + (AF + Maxdigs - 1) / Maxdigs;
-- Number of rounds of scaled divide to be performed
Q : Int;
-- Quotient of the scaled divide in this round. Only the first round may
-- yield more than Maxdigs digits and has a significant sign.
Buf : String (1 .. Maxdigs);
Len : Natural;
-- Buffer for the image of the quotient
Digs : String (1 .. 2 + N * Maxdigs);
Ndigs : Natural;
-- Concatenated image of the successive quotients
Scale : Integer := 0;
-- Exponent such that the result is Digs (1 .. NDigs) * 10**(-Scale)
XX : Int := V;
YY : Int := Y;
-- First two operands of the scaled divide
begin
-- Set the first character like Image
if V >= 0 then
Digs (1) := ' ';
Ndigs := 1;
else
Ndigs := 0;
end if;
for J in 1 .. N loop
exit when XX = 0;
Scaled_Divide (XX, YY, Z, Q, R => XX, Round => False);
if J = 1 then
if Q /= 0 then
Set_Image_Integer (Q, Digs, Ndigs);
end if;
Scale := Scale + D;
-- Prepare for next round, if any
YY := 10**Maxdigs;
else
pragma Assert (-10**Maxdigs < Q and then Q < 10**Maxdigs);
Len := 0;
Set_Image_Integer (abs Q, Buf, Len);
pragma Assert (1 <= Len and then Len <= Maxdigs);
-- If no character but the space has been written, write the
-- minus if need be, since Set_Image_Integer did not do it.
if Ndigs <= 1 then
if Q /= 0 then
if Ndigs = 0 then
Digs (1) := '-';
end if;
Digs (2 .. Len + 1) := Buf (1 .. Len);
Ndigs := Len + 1;
end if;
-- Or else pad the output with zeroes up to Maxdigs
else
for K in 1 .. Maxdigs - Len loop
Digs (Ndigs + K) := '0';
end loop;
for K in 1 .. Len loop
Digs (Ndigs + Maxdigs - Len + K) := Buf (K);
end loop;
Ndigs := Ndigs + Maxdigs;
end if;
Scale := Scale + Maxdigs;
end if;
end loop;
-- If no digit was output, this is zero
if Ndigs <= 1 then
Digs (1 .. 2) := " 0";
Ndigs := 2;
end if;
Set_Decimal_Digits (Digs, Ndigs, S, P, Scale, Fore, Aft, Exp);
end Set_Image_Fixed;
end System.Image_F;