Please explain in detail how the answers above were found Th

Please explain in detail how the answers above were found. Thank You!

For each of the following program segments, execute the program one instruction at a time. At the end of each instruction execution, indicate any changes in registers, condition flags, and or memory locations. Once you compile the program segments below and before starting the execution on one instruction at a time by using the CCS step-over function, you need to manually fill in the initial data in the registers and memory. Note that the result of any instruction may be an operand for succeeding instructions.

Solution

addc.w @r5+, r6       \\\\ contents present at memory location 0208 is 5500 + a006 gives f507 which is stored in r6 and r5 becomes 20a i.e next memory location after increment.(@r5+).

mov &0204h, r6        \\\\ move the content 3300h which is present in the memory location 204h to r6.

mov.w r6, 0(r5)         \\\\ the content present in r6 i.e 3300h is moved into the memory contents of the location of r5(which is displayed in cell of last row last column).

bic.w #0x3100, r6      \\\\ bic operates as \"and not\" i.e. this instruction performs an AND operation on the bits with the complements of the corresponding bits in the value of operand2 i.e. r6 and the value gets stored in r6.

0011 0011 0000 0000

1100 1110 1111 1111 (compliment of 3100)

= 0000 0010 0000 0000 = 0200h.

subc 6(r6), -2(r5)       \\\\ subtract and carry operator will subtract the (200+6) -(- 20a+2)= 206 -(- 20c) = 206 - (compliment the value 20c in binary) = 10ff which is stored in memory contents of 208 location since -2(r5) means decrement the location also 2 times and since carry is obtained value under C column is 1.