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 

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 !!! 

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 !) 
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) 