Re: Numeric problem with Ada

[CNAM.CNAM.FR!bortz@ucbvax.Berkeley.EDU]

Thanks to all for the replies to my question about why C and Ada give
two different results in "equivalent" numeric programs.
Two kinds of solution were presented: those who said it was a problem
in the I/O library and those who were looking toward the use of
double-precision arithmetic by the C compiler. From the following
modification of my test program (to suppress problems related with
I/O), I think the second hypothesis is the good one.
Here is the new program:
Ada:
	if Z2 /= 18.0 then 
              TEXT_IO.PUT_LINE ("Z2 /= 18.0"); 
        else 
              TEXT_IO.PUT_LINE ("Z2 = 18.0"); 
        end if;
C:
	if (z2 != 18.0) printf ("z2 != 18.0\n"); else printf ("z2 = 18.0\n"); 

Ada finds that Z2 /= 18.0 and C that z2 = 18.0. 
So the difference is really in internal computations. Two replies by
E-mail were specially interesting:

pragma BEGIN_QUOTATION (hilfingr@tully.cs.berkeley.edu "Paul N. Hilfinger");
C implementations generally DO use double precision (the new ISO stanard
calls for it in this case; both of the constants 2.6 and 0.2 are
double-precision, lacking any suffix to the contrary).  When I tried
this on a Sun 3, the compiler generated double-precision instructions
and got the result you indicated (18.000000).  When I changed the
calculation of z2 to read

       z2 = (float) 2.6 * mz - (float) 0.2;

the result was printed as 17.999999 (I used gcc version 1.39 on a
Sun3).  Perhaps you misread the assembly language code (those mnemonics
ARE rather cryptic). 
pragma END_QUOTATION (hilfingr@tully.cs.berkeley.edu "Paul N. Hilfinger");

I agree with the possibility that I missed something in the assembly
code (I prefer to read Ada!).

pragma BEGIN_QUOTATION (Per-Gunnar.Johansson@imsa.hv.se "Per-Gunnar Johansson")
;
The computer store Your object in a binary representation, and if You have
six digits in the decimal system You have about 20 digits in the binary
system.
If You wamt to store a float-object the coputer transform it on the form
    number = m * (2 ** e), where 0.5 <= m < 1 or m = 0.
It is no problem with the number 7 (7 = 0.875 * (2 ** 3), because 
0.875 in the decimal system is exact 0.111 in the binary system.
BUT, the number 2.6 = 0.65 * (2 ** 2), and 0.65 in the decimal system
is 0.1010 0110 0110 0110 0110 ..... (and so on), in the binary system.
The representation in Your computer, with six decimal digits, is about 20
binary digits. It means that Your computer represent the number 2.6 with
m = 0.1010 0110 0110 0110 in the binary systemand it is about 0.649999619
in the decimal system, so (2.6 =) 0.649999619 * (2 ** 2) = 2.599998476.
In the same way (0.2 =) 0.199999809
In that way, You calculating operation
    2.6 * 7 - 0.2 =
  = 20599998476 * 7 - 0.199999809 =
  = 17.99998952
It is not 17.999998 but I think it maybe depend on the accurancy on my
pocket calculator.
pragma END_QUOTATION (Per-Gunnar.Johansson@imsa.hv.se "Per-Gunnar Johansson");


Stephane Bortzmeyer           Conservatoire National des Arts et Metiers
	
bortzmeyer@vcnam.cnam.fr      Laboratoire d'Informatique
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