(Why) is using an uninitialized variable undefined behavior?

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Yes this behavior is undefined but for different reasons than most people are aware of.

First, using an unitialized value is by itself not undefined behavior, but the value is simply indeterminate. Accessing this then is UB if the value happens to be a trap representation for the type. Unsigned types rarely have trap representations, so you would be relatively safe on that side.

What makes the behavior undefined is an additional property of your variable, namely that it “could have been declared with register” that is its address is never taken. Such variables are treated specially because there are architectures that have real CPU registers that have a sort of extra state that is “uninitialized” and that doesn’t correspond to a value in the type domain.

Edit: The relevant phrase of the standard is 6.3.2.1p2:

If the lvalue designates an object of automatic storage duration that
could have been declared with the register storage class (never had
its address taken), and that object is uninitialized (not declared
with an initializer and no assignment to it has been performed prior
to use), the behavior is undefined.

And to make it clearer, the following code is legal under all circumstances:

unsigned char a, b;
memcpy(&a, &b, 1);
a -= a;
  • Here the addresses of a and b are taken, so their value is just
    indeterminate.
  • Since unsigned char never has trap representations
    that indeterminate value is just unspecified, any value of unsigned char could
    happen.
  • At the end a must hold the value 0.

Edit2: a and b have unspecified values:

3.19.3 unspecified value
valid value of the relevant type where this International Standard imposes no requirements on which value
is chosen in any instance

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