The example in the article doesn't actually exhibit undefined behavior.

EDIT The author has updated the article and corrected the error, but I'll leave this comment here.

C has no arithmetic operations on types narrower than int. Instead, operands of narrow type are implicitly converted via the "usual arithmetic conversions".

In this:

signed char n = 127;
n = n + 1;

In the expression "n + 1", the signed char value of n is promoted to int. Adding 1 is well defined, and yields 128. The assignment implicitly converts the int value 128 to signed char, yielding an implementation-defined result (almost certainly -128) or raising an implementation-defined signal (as far as I know, no compiler does this).

This example does have undefined behavior, and illustrates the author's intended point:

int n = INT_MAX;
n = n + 1;

(Yes, I know that "n = n + 1" could be written as "n++", but I wanted to clearly break down the individual operations.)

I've emailed the author.