Chapter 6 Applied Instructions Example 1 1. When M0 is ON, the instruction converts the binary integer in D0 into a single-precision floating-point number, and stores the conversion result in (D13, D12). 2. When M1 is ON, the instruction converts the binary integer in (D1, D0) into a single-precision floating-point number, and stores the conversion result in (D21, D20). 3. Suppose the value in D0 is 10. When M0 is ON, the instruction converts 10 into the single-precision floating-point number 16#41200000, and then stores 16#41200000 in the 32-bit register (D13, D12). 4. Suppose the value in the 32-bit register (D1, D0) is 100,000.
When M1 is ON, the instruction converts 100,000 into the single-precision floating-point number 16#47C35000, and stores 16#47C35000 in the 32-bit register (D21, D20). 6_ Example 2 You can use the applied instructions to perform the following calculation. Convert the binary integer in D10 into the single-precision floating-point number, and store the conversion result in (D101, D100). Move the binary-coded decimal value in X0–X17 to D150. Convert the binary-coded decimal value in D150 into the binary value and store the conversion result in D150. Convert the binary integer in D200 into the single-precision floating-point number, and store the conversion result in (D203, D202).
Divide the constant 615 by the constant 10, and store the quotient which is the single-precision floating-point number in (D301, D300). Divide the single-precision floating-point number in (D101, D100) by the single-precision floating-point number in (D203, D202), and store the quotient which is the single-precision floating-point number in (D401, D400). 6-85