3 . I ns t r uc ti on Se t3- 3 1 7Program Example 1:When X0 is ON, the dividend K103 in D0 is divided by the divisor K5 in D10, and the quotient isstored in D20. Whether the quotient is a positive value or a negative value depends on the leftmostbit in D20.X0DI V16 D0 D10 D20D0/D10=D20 K103/K5=K20, the remainder is K3. D20=K20 (The remainder is left out.)Program Example 2:When X0 is ON, the dividend K81,000 in (D1, D0) is divided by the divisor K40,000 in (D11, D10),and the quotient is stored in (D21, D20). Whether the quotient is a positive value or a negative valuedepends on the leftmost bit in (D21, D20).X0DI V32 D0 D10 D20(D1,D0)/(D11,D10)=(D21,D20) K81,000/K40,000=K2, The remainder is K1,000. (D21,D20)=K2 (The remainder is left out.)Note:1. If users want to store the remainder of a 16-bit bianry division, they have to use API23 DIV/DIVP.Please refer to the explanation of API23 DIV/DIVP for more information.2. If users want to store the remainder of a 32-bit bianry division, they have to use API23DDIV/DDIVP. Please refer to the explanation of API23 DDIV/DDIVP for more information.
