Implementing High Level Integer Arithmetic Operations 1

1
Implementing High Level Integer Arithmetic
Operations - 1

• The Basic Issues
• Given a specified n bit CPU ( n8 /16 /32 ).
• Perform the following arithmetic operations
• Add / Subtract two m digit integers ( unsigned /
  signed) A B in any base to produce the m digit
  (unsigned/signed) result C in that base .
• Multiply two unsigned / signed m digit integers
  A B in any base producing a 2m digit unsigned /
  signed integer result C.
• Divide a 2m digit integer dividend
  (unsigned\signed) C by another m digit integer
  divisor (unsigned/signed) B to produce an m digit
  integer (unsigned/signed) quotient Q as well as
  another m digit integer (unsigned/signed)
  remainder R.

2
Implementing High Level Integer Arithmetic
Operations - 2

• The Facts the associated assumptions
• The Fact An m digit integer (signed/unsigned),
  in any base when converted to binary yields a k
  bit binary integer (signed/unsigned) where k gt
  n and k need not be an integer multiple of n.
• The Assumptions
• All Operands A , B OR C as well as the Results
  ( C , Q R) are expressed as integer multiples
  of n. ( The CPU Size) and all of them are stored
  in memory. Each is expressed in HEX.
• The CPU size ( n ) is a power of 2 I.e. n 2 j
  where j has one of the following integer values
  2 4 bit CPU , 3 8 bit CPU , 4 16 bit CPU
  or 5 32 bit CPU. We assume here j 5 I.e. n
  32.
• Memory is having 32 bit address, speed compatible
  with CPU, alterable and byte organized.
• Everything in memory is stored in n bit word
  boundary in Little Endian convention, in
  consecutive locations starting from a000 0000H

3
The Memory Storage Pattern

• Assume A 3e 4b 6c bf 67 79 88 ae 7b ee fd ae ff
  b8 a9 bf Hex 16 bytes
• B f4 56 ab 56 78 99 8b 6e de fa
  88 98 ac Hex 13 bytes
• ff ff ff f4 56 ab 56 78 99
  8b 6e de fa 88 98 ac Hex 16 bytes
• After sign Extension
• C A op B 1d 2e .. .. .. .. .. .. bf 67 bf
  de ae ff Hex 16 byte Result
• Variable Memory Offset
  Content
• --------------------------------------------------
  ------------
• A a000 0000H
  a9
• --------------------------------------------------
  ------------
• A1 a000 0001 H
  b8
• .. A15 a000 000f H
  3e
• --------------------------------------------------
  ------------
• .. B a000 0010 H
  ac
• B15 a000 001f H
  ff
• --------------------------------------------------
  -----------
• C a000 0020 H
  ff
• .. C15 a000 002f H
  1d

4
Our Assumptions Contd.

• A 32 bit CPU with 32 bit Data Bus 32 bit
  Address Bus.
• 8 Number of 32 bit GPRs r0 .. r7.
• All operands must be brought inside some of the
  GPRs of the CPU before any processing.
• The 32 bit ALU is involved in all types of
  processing.
• All results (intermediate as well as final) are
  first generated and stored inside some GPR of the
  CPU.

5
The Addition / Subtraction Operation

• Task at hand
• Given Operand A 128 bit long stored from
  location a000 0000H.
• Operand B 128 bit long stored from location
  a000 0010 H.
• To Compute C A PLUS B OR A MINUS B with C to be
  stored from location a000 0020 H.
• All values are stored using little Endian
  conventions.

6
The Addition Operation Steps

• Set up Word Counter . r5 ? 4
• Clear CARRY Flag.
• Set up address pointer for Operand A r0 ?
  Addr. of A .
• Set up address pointer for Operand B r1 ?
  Addr. of B .
• Set up address pointer for Operand C r2 ?
  Addr. of C .
• Load a GPR with the Current Word of Opnd. A r6
  ? M(r0) .
• Load a GPR with the Current Word of Opnd. B r7
  ? M(r1).
• Add with Carry r6 ? (r6) PLUS (Sr7). PLUS
  (Cy). Ignore Overflow.
• Store Result in Current Word of Operand C
  M(r2) ? r6.
• Increment Address Pointers r0 (A) , r1(B),
  r2(C) .
• Decrement Word Counter r5.
• If Word Counter 0 GOTO EXIT ELSE GOTO Step 6.
• EXIT Check Overflow Generate App. Message
  GOTO Monitor.

7
The Subtraction Operation Steps

• Set up Word Counter . r5 ? 4
• Clear CARRY/ BORROW Flag.
• Set up address pointer for Operand A r0 ?
  Addr. of A .
• Set up address pointer for Operand B r1 ?
  Addr. of B .
• Set up address pointer for Operand C r2 ?
  Addr. of C .
• Load a GPR with the Current Word of Opnd. A r6
  ? M(r0) .
• Load a GPR with the Current Word of Opnd. B r7
  ? M(r2).
• Invert CARRY / BORROW Flag,
• Subtract with Borrow r6 ? (r6) MINUS (Sr7)
  MINUS Cy. Ignore Overflow.
• Store Result in Current Word of Operand C
  M(r2) ? r6.
• Increment Address Pointers r0 (A) , r1(B),
  r2(C) .
• Decrement Word Counter r5.
• If Word Counter 0 GOTO EXIT ELSE GOTO Step 5.
• EXIT Check Overflow Generate App. Message
  GOTO Monitor.

8
Multiplication

• Following two ( 2 ) types of operations
• 1) Unsigned Multiplication
• C A X B where A B each happens to be
  unsigned n digit operands C is the unsigned 2n
  digit result.
• 2) Signed Multiplication
• C A X B where A B each happens to be signed
  
• ( rs complement) n digit operands C is the
  signed (rs complement) 2n digit result. In each
  case the MS Digit stores the sign.

9
Division

• Following two ( 2 ) types of operations
• 1) Unsigned Division
• C / A yielding Q R such that C Q X A R
• where C is the unsigned 2n digit Dividend.
• A is the unsigned n digit divisor.
• Q is the unsigned n digit quotient.
• R is the unsigned n digit remainder.
• 2) Signed Division
• C / A yielding Q R such that C Q X A R
• where C is the signed 2n digit Dividend.
• A is the signed(rs complement) n digit
  divisor.
• Q is the signed (rs complement) n digit
  quotient.
• R is the signed (rs complement) n digit
  remainder ( same sign as C).