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The MULS instruction - know when to use ext.l to convert into long

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• RECALL: Mulitply instruction in M68000
  • M68000 can only multiply two 16-bits integers (due to the technological limitation at the time - 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 result is always 32 bits and it is stored in data register Dn

  • Notice that you do not have any choice for operand size.

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• Multiple int data types
  • Facts:

    int data type uses 32 bytes to represent the value When an int is used in the muls instruction: The 32 bit representation is truncated to a 16 bit representation

  • Example:

    int i1, i2, i3; i3 = i1 * i2; In assembler code: move.l i1,D0 * get 32 bits value i1 in reg D0 move.l i2,D1 * get 32 bits value i2 in reg D1 muls D1,D0 * D0 = D0*D1 * We only use 16 bits in D0 and D1 * So we have converted the int * into a short before we multiply !! * Note: result is correct as long as * the values in D0 and D1 is small move.l D0,i3 * Store i1*i2 to i3 * The product is 32 bits !!!

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• Multiply with byte size operands
  • Important fact:

    MULS will always use 16 bits operands !!! So we must convert a byte (8 bits) representation into a 16 bit representation before we use MULS !!!

    Example:

    byte b1, b2, b3; b3 = b1 * b2; In assembler code: MOVE.B b1, D0 * D0 = b1 (8 bits) EXT.W D0 * D0 now has a 16 bits representation !! MOVE.B b2, D1 * D1 = b2 (8 bits) EXT.W D1 * D1 now has a 16 bits representation !! MULS D1, D0 * D0 = b1 * b2 (32 bits) * The product is 32 bits !!! MOVE.B D0, b3 * Move byte value to b3 (We have actually converted an int into a byte !)

  • Try out this demo program yourself: click here

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• Some more examples
  • More examples:

    int a; short b; byte c; a = b * c; move.w b, d0 (16 bits valid in d0) move.b c, d1 (8 bits valid in d1) ext.w d1 (16 bits valid in d1) * Now have two 16 bits values and can use muls !! muls d0, d1 (32 bit result in d1) move.l d1, a (store 32 bits in a) b = a * c; move.l a, d0 (32 bits valid in d0) (We will only use 16 bits) move.b c, d1 (8 bits valid in d1) ext.w d1 (16 bits valid in d1) * Now have two 16 bits values and can use muls !! muls d0, d1 (32 bit result in d1) (We will only use 16 bits) move.w d1, b (store 16 bits in b) c = a * b; move.l a, d0 (32 bits valid in d0) (We will only use 16 bits) move.w c, d1 (16 bits valid in d1) * muls will only use 16 bits from d0 muls d0, d1 (32 bit result in d1) (We will only use 8 bits) move.b d1, c (store 8 bits in c)