Pfthb

The question confuses me a long time. How to detect overflow in such situation which 0 subtracts (-128) in
8 bits environment.

In class, my teacher teach me a method
to detect overflow, the method is as below:

But it doesn't work in such situation:

Above situation computer can detect overflow. But I don't know how computer does it.

You've got an error in your calculation.

Subtracting two numbers in two's complement is not how you've depicted it in the second image.

The image shows addition, not subtraction, and in subtraction you use borrow not carry.

The borrow is where you've made your mistake.

 10000000 (borrow)
 00000000 (0)
- 10000000 (-128)
----------
 10000000 (-128)

As you can see the first two bits of the borrow are different hence 1 XOR 0 = 1 yields overflow.

See Wikipedia for more information.

Addendum

To clarify why your assumption of that an overflow has not occurred, when infact it has, in your addition calculation in the second image, is wrong:

 00000000 (carry)
 00000000 (0)
+ 10000000 (-128)
----------
 10000000 (-128)

Since 0 + -128 = -128 and 0 XOR 0 = 0 hence no overflow.

In this case the addition doesn't cross the -128 and 127 boundary.

Let's look at a representation of two's complement with only 4-bits.

-128 would be represented by -8.

If you go from 0 to -8 counter-clockwise (for subtraction) you'll see that the boundary (depicted by the red line) is not crossed, hence no overflow.

If we take your first example which would be 7 + 1 and go clockwise (addition) from 7 you'll end up at -8 and you have crossed the boundary, hence overflowed.

What you thought you did in your second image was 0 + 128. You went from 0 to -128 clockwise (addition). This crossed the boundary but since 128 cannot be represented in 8-bit two's complement the assumption and calculation was wrong.

You assumed 0 + 128 was the same as 0 - (-128) which it clearly isn't since as said above 128 cannot be represented in 8-bit two's complement.