Q60037: Floating-Point Differences Between the 8086 and MC68000

Article: Q60037
Product(s): See article
Version(s): 5.10   | 5.10
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | s_quickc s_quickasm h_masm endian | mspl13_c
Last Modified: 15-JAN-1991

Question:

I am writing a file that will be exchanged between an MC68000 machine
and an 8086 machine. I want to know how the byte ordering of the IEEE
(Institute of Electrical and Electronic Engineers) floating-point
format must be manipulated. In other words, how should the bits or
bytes be reordered?

Response:

The byte ordering of floating point numbers in the architecture of the
MC68000 is opposite that of the 8086; that is, the bytes are reversed.
Microsoft C Version 5.10 conforms to the IEEE standard. The 8086
stores floating-point numbers low-order byte to high-order byte (known
as "little end in"). The MC68000 stores floating-point numbers
high-order byte to low-order byte (known as "big end in"). Except for
the bytes being reversed, the internal representation of the float is
the same.

For example, given the following floating-point number in the
following 4-byte real data type

   5.27E-3

the representation of that number in IEEE format would be

   3B AC AF F7

which is equivalent in binary to the following:

   0011 1011 1010 1100 1010 1111 1111 0111

However, an Intel machine stores this number with the bytes reversed,
as follows:

   F7 AF AC 3B

On the other hand, in MC68000 format, the bytes are not reversed.
Internally the MC68000 would represent the above number as

   3B AC AF F7

which is the same as the IEEE representation; therefore, no "manual"
byte reversal needs to be effected.

To summarize the internal interpretation of the above number, the
following information is valid for both IEEE and MC68000 formats:

The mantissa is bits 0-22, which in this case equals 2928631. The
mantissa is assumed to be prefixed with an implied "1."; therefore,
the full mantissa is 1.2928631. Bits 23-30 represent the exponent.
This exponent must be adjusted by a factor of 127 (80 hex; the "bias"
adjustment, which must be subtracted from the bitwise representation
of the exponent). Bit 31 is 0, which implies that the exponent is
positive. Therefore, the exponent as represented here is 119.
Performing the following bias subtraction:

   119-127 = -8

While the base in the original number 5.27E-3 is 10, the base of the
internal representation is 2. Therefore, the above calculation yields
the final result of

   1.2928631 X 2E-8

which is the original number (allowing for some floating point
inaccuracies).

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