Question

The following code calculates A = (4 * B) + C in which A, B, and C are signed bytes. If C has the decimal value of +28, then what is the largest decimal value for B that allows the code to correctly calculate A = (4 * B) + C as a signed byte?

mov al, [B] ; B

add al, [B] ; 2 * B

add al, al ; 4 * B

add al, [C] ; (4 * B) + C

move [A], al ; A = (4 * B) + C

A) +15 B) +19 C) +24 D) +28 E) +31

Could someone please explain this problem to me as they provide the solution? Thank you.

Answer 1

A = (4 * B) + C

First signed byte range from -128 to 127.
so whatever the value of A it should be be in that range.

so let's put all the values B and find the result.

A) A = (4 * 15) + 28 = 88
B) A = (4 * 19) + 28 = 104
C) A = (4 * 24) + 28 = 124
D) A = (4 * 28) + 28 = 140
E) A = (4 * 31) + 28 = 152

D,E are above the range of signed byte
B,A are below the range and the C
C is the largest possible value for a signed byte. so answer is C (B = 24).
The value for C is 28.