Advanced Arithmetic

1
Advanced Arithmetic

• Assembly Language Programming
• University of Akron
• Dr. Tim Margush

2
Double-Precision Integers

• The 8086 integer operations are limited to
  16-bits
• If larger numbers are required (32, 64 bits), how
  do we handle the arithmetic?
• Emulate the arithmetic operations
• Provide additional hardware to perform arithmetic

3
32-Bit Emulation Add

• To add two 32-bit numbers in memory
• x dd 12345678h
• y dd 98765432h
• z dd ? result
• Add low order words and store result
• Remember they are stored in reverse byte format

• mov ax,word ptr x
• add ax,word ptr y
• mov wordptr z,ax
• The carry must be added to the sum of the
  high-order words
• mov ax,word ptr x2
• adc ax,word ptr y2
• mov word ptr z2,ax

4
32-Bit Emulation Sub

• To subtract two 32-bit numbers in memory
• x dd 12345678h
• y dd 98765432h
• z dd ? result
• Sub low order words and store result
• Remember they are stored in reverse byte format

• mov ax,word ptr x
• sub ax,word ptr y
• mov wordptr z,ax
• The carry (borrow) must be added to the sum of
  the high-order words
• mov ax,word ptr x2
• sbb ax,word ptr y2
• mov word ptr z2,ax

5
Flags

• The flags will not reflect the result of the
  double-precision operation
• SF, OF, and CF are correct after the second
  operation
• If necessary, set the other flags manually
• ZF is set if both words are zero
• PF depends on parity of both words
• AF is probably unnecessary to correct

6
Multiplication

• 12345678h 98765432h
• 5678h 5432h 1C704370h
• 1234h 5432h 5FC9E28h
• 5678h 9876h 337F1B50h
• 1234h 9876h AD743F8h

• 1C70 4370
• 5FC 9E28
• 337F 1B50
• AD7 43F8
• ----------------
• 0AD7 7D73 D5E8 4370
• If signed numbers are involved, suitable
  conversions should be performed

7
Converting Decimal Fractions

• decimal fraction .ddddd to binary .bbbbb
• Successive doubling of the fraction tells whether
  the next bit of the binary representation is 1 or
  0
• if f 2gt1 then next bit is 1
• if f 2 lt 1 then next bit is 0
• We discard the whole part before the next doubling

8
Example Conversion

• .24 2 0.48
• .48 2 0.96
• .96 2 1.92
• .92 2 1.84
• .84 2 1.68
• .68 2 1.36
• .36 2 0.72
• .72 2 1.44

• .44 2 0.88
• .88 2 1.76
• .76 2 1.52
• .52 2 1.04
• the process continues for as many bits as are
  desired
• .001111010111...

9
Floating Point Representation

• Mantissa
• Significant digits of the number
• Normalized if 1.bbbbbbbb...
• Exponent
• Specifies adjustment to the location of decimal
  (binary) point
• Example 101.1101 with exponent 2 is the same as
  101110.1 with exponent -1

10
Encoding Floating Point Numbers

• Sign and Magnitude
• One bit in the code represents the sign of the
  number
• The magnitude (expressed in normalized form) is
  encoded in the remaining bits
• Exponent and Mantissa
• The Exponent is translated using a biased code
  and stored in a field of the code
• The Mantissa bits are stored in the remaining
  bits of the code

11
Short Real
s
exponent
mantissa

• 32-bit floating point code
• bit 31 is the sign bit (1 negative)
• bits 30-23 hold the exponent
• Exponents are biased by 127
• Exponent range is -126 to 127
• bits 22-0 the mantissa
• The leading 1 is not stored.

12
Example Floating Point Conversion

• 42.25 (decimal) converts to 101010.01b
• Normalize 1.0101001 exponent 5
• sign bit 0
• exponent 10000100b (5127132)
• mantissa 01010010000000000000000
• leading 1 is dropped
• 0100 0010 0010 1001 0000 0000 0000 0000
• Hexadecimal 42290000h

13
Decoding Short Reals

• 4372F10Ch
• Assuming this is a Short Real, what decimal
  number does it represent?
• Decode first 9 bits
• 0 10000110
• sign bit 0

• Exponent code is 134
• 134-1277 (exponent)
• Mantissa is 1.11100101111000100001100
• Move the binary point
• 11110010.1111000100001100
• 242.9415893555...