CODEWITHSUNDEEP
cpu-architecture
Teradice
Teradice

Reputation: 53

Is this a mistake in my Computer Architecture book?

I'm working on my HW for computer architecture and I came across the following problem:

A = 247
B = 237

1) Assume A and B are signed 8-bit integers stored in two's complement format. Calculate A + B using saturating arithmetic. The result should be written in decimal. Show your work.
2) Assume A and B are signed 8 bit integers stored in two's compelemnt format. Calculate A - B using saturating arithmetic.

Now, how are these even a valid questions? I know what saturating arithmetic is, but how is it valid to say that A is 247 and B is 237 when they can't be represented by an 8-bit two's complement number?

I realize the point of saturated arithmetic is in the case of a overflow/underflow to set all the bits to the extreme value but it doesn't make sense to me to ask a series of questions (there are 3 more with this same problem) involving arithmetic of numbers that can't be represented in the format they specify.

Am I wrong here?

Upvotes: 5

Views: 2728

Answers (4)

gnasher729
gnasher729

Reputation: 52538

If the problem was "assume that A and B are stored in an 8 bit signed integer", then it would make sense. Storing A = 247 in a signed 8 bit integer yields a value of -9, storing B = 237 yields a value of -19. Otherwise, it doesn't make sense. A signed 8 bit integer cannot have a value of 247 or 237.

Upvotes: 0

poundifdef
poundifdef

Reputation: 19353

I don't know, but it might be asking: "Convert the decimal number 237 to an 8-bit integer. Now interpret those bits as an 8-bit 2's complement integer and add them".

Kinda like saying, in C:

char a = 237;
printf("%x %d\n",a, a);

Which compiles and gives you values that you would expect based on the 2's complement interpretation of the 8-bit value "237"

Upvotes: 3

bdonlan
bdonlan

Reputation: 231103

The only interpretation that makes sense is that the values given are the unsigned interpretation of the number in question; values greater than 127 are obviously out of range for an 8-bit signed twos-complement value. I agree that the question is poorly stated, however.

Upvotes: 8

Aric TenEyck
Aric TenEyck

Reputation: 8032

Looks wrong to me. 8-bit signed two's complement integers can be between -128 and 127. Any attempt to assign a constant 237 or 247 to an 8-bit signed two's complement variable should result in an error, depending on your language.

Upvotes: 1

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