Chapter 6, Problem 42E

Explanation of Solution

Given code:

The given program is as follows:

                BR         main

sum:      .WORD      0x0000

num1:     .BLOCK     1

num2:     .BLOCK     1

num3:     .BLOCK     1

main:     LDWA     sum,d

          DECI     num1,d

          DECI     num2,d

          DECI     num3,d

          ADDA     num3,d

          ADDA     num2,d

              ADDA     num1,d

          STWA      sum,d

          DECO     sum,d

          STOP

          .END

Explanation:

• The above code is to get the input for the variables “num1”, “num2”, and “num3” from the user.
• Then, the sum of three numbers is calculated and the total value on the screen is displayed.
• The highlighted lines in the code shows that the storage setup for the input variables “num1”, “num2”, and “num3” is only 1 byte.
• If the size of the input value is greater than 1 byte, then the given code results in a wrong output.
  • As per the test plan used in this chapter, the largest input value is 32767; it needs two bytes of memory to store the value.
  • Therefore, wrong results can be avoided by declaring the variables with two bytes.

Corrected code:

The bug in the given code is rectified and the corrected code is as follows:

                BR         main

sum:      ...