I understand the rule says that overflow cannot happen for two's complement addition of numbers with different signs, but do not understand why this specific case does not cause overflow:
1001010
+0010101
--------
1011111
Am I simply misunderstanding the actual rule or missing a step somewhere? Thanks in advance!
EDIT: removed an extra 1 in the result
Okay, let's take a look at your example:
1001010
+0010101
--------
1011111
Your numbers here only have 7 bits, so we have a smaller range than we typically deal with. Nevertheless:
$1001010$ is
$-64 + 8 + 2$, which is $-54$.
$0010101$ is
$16 + 4 + 1$, which is $21$.
$21-54$ is $-33$. Let's see whether our answer matches up:
$1011111$ is
$-64+16+8+4+2+1$, which is $-33$.
An intuitive way to think about this is that the number space is divided entirely in half (with $0$ acting as a positive number). So, with the standard 8 bits, your numbers range from $-128$ to $127$. There is no positive number from within that set that you could add to a negative number from that set and get a number either above $127$ or below $-128$.
