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The MULS and DIVS instructions

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• Precision consideration in multiplication and division
  • Fact about the result of a addition and/or subtraction:

    The absolute value of the sum of 2 numbers a + b is at most twice the absolute value value of the largest number Example: 10 + 9 = 19 (≤ 2 × 10) The absolute value of the difference of 2 numbers a - b is at most twice the absolute value value of the largest number Example: 10 - (-9) = 19 (≤ 2 × 10)

    Notice that:

    The number of digits of the sum is approximately the same as the source numbers a and b !!!

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  • On the other hand:

    The absolute value of the product of 2 numbers a × b can be much large than either number (a or b) Example: 1000 × 1000 = 1000000 1000 has 4 digits 1000000 has 7 digits ! The absolute value of the quotient of 2 numbers a / b can be much smaller than either number (a or b) Example: 100000 × 100000 = 1 100000 has 6 digits 1 has 1 digit !

    Therefore:

    To preserve accuracy, the data type of the result of the multiply (×) operation is increased

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• Mulitply instruction in M68000
  • Warning:

    The M68000 can only multiply two 16-bits integer numbers (This limitation is due to the technological limitation at the time - circa 1980)

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  • The syntax of the multiply instruction of M68000 is:

    MULS <ea>, Dn Multiply the 16 bit integer value in the operand specified by <ea> to the 16 bit value in data register Dn The product is always 32 bits and it is stored in data register Dn In other words: Dn(16 bits) × <ea>(16 bits) ⇒ Dn(32 bits)

  • Note:

    You do not have any choice for operand type in the MULT instruction: You must use a short (16 bits) values to multiply Also, one of the integer (number) that you multiply must be stored inside a data register

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• Example of the MULS instruction
  • Suppose you have the following bit pattern in D0:

    +----------+----------+----------+----------+ D0 = | 10101010 | 01010101 | 00000000 | 00001001 | +----------+----------+----------+----------+ The 16 bit operand in D0 is equal to: (00000000 00001001)(2) = 1 + 8 = 9(10)

    The 16 bit representation in register D0 represents the decimal value 9₍₁₀₎

    Suppose we execute the following multiply instruction:

    MULS #3, D0

    The data register D0 will contain the following result after the instruction is executed:

    +----------+----------+----------+----------+ D0 = | 00000000 | 00000000 | 00000000 | 00011011 | +----------+----------+----------+----------+ The 32 bit result in D0 is equal to: (00000000 00000000 00000000 00011011)(2) = 1 + 2 + 8 + 16 = 27(10)

    Notice that all 32 bits in the register D0 are updated because the product of two 16-bit binary number is always 32 bits.

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• Divide instruction in M68000
  • Important note:

    The M68000 can only divide a 32-bits integer number by a 16-bits integer number (This, again, is due to the technological limitation at the time in 1980)

  • The syntax of the DIVS instruction is:

    DIVS <ea>, Dn Divides a 32 bit value in data register Dn by a 16-bit value specified by <ea> In other words: Dn(32 bits) / <ea>(16 bits) Result: (1) the quotient is stored in the lower 16 bits of data register Dn (2) the remainder is stored in the upper 16 bits of data register Dn

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    • Warning:

      the DIVS instruction can result in error !!! E.g.: 1 / 0 !!!! The DIVS is voided if the execution results in error

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• Example of the DIVS instruction
  • Suppose register D0 contains the value 9 (in binary !):

    +----------+----------+----------+----------+ D0 = | 00000000 | 00000000 | 00000000 | 00001001 | +----------+----------+----------+----------+ The 32 bits represents the value 9(10)

    Then after executing the following DIVS instruction:

    DIVS #4, D0 9 / 4 --> Quotient = 2 Remainder = 1

    The data register D0 will contain:

    +----------+----------+----------+----------+ D0 = | 00000000 | 00000001 | 00000000 | 00000010 | +----------+----------+----------+----------+

    • The quotient is stored in the lower half of the register
    • The remainder is stored in the upper half of the register

  • NOTE: when you use EGTAPI and display register D0, you need to display it in half word quantity to see the quotient and remainder parts !

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• The SWAP instruction
  • Notice that:

    the remainder of a division is stored in the upper half of the data register.

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  • However:

    Word size operands are stored in the lower half in the data registers

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  • Fact:

    The SWAP instruction is used to: make the remainder of the division available as operand in a data register

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  • The SWAP instruction:

    The SWAP Dn instruction exchanges the upper and the lower halves of the data register Dn.

  • Example:

    Suppose data register D0 contains the following: +----------+----------+----------+----------+ D0 = | 00000000 | 00000001 | 00000000 | 00000010 | +----------+----------+----------+----------+ After executing the following unstruction: SWAP D0 (= exchanges the upper and lower halves of D0) The data register D0 will contain the following: +----------+----------+----------+----------+ D0 = | 00000000 | 00000010 | 00000000 | 00000001 | +----------+----------+----------+----------+

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• Teaching notes
  • We will learn how to use the MULS, DIVS and SWAP instructions later

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  • We need to cover:

    Converting operand sizes first !!!