(Why) is using an uninitialized variable undefined behavior?

Question

If I have:

unsigned int x;
x -= x;

it\'s clear that x should be zero after this expression, but everywhere I look, th

Answer 1

(This answer addresses C 1999. For C 2011, see Jens Gustedt’s answer.)

The C standard does not say that using the value of an object of automatic storage duration that is not initialized is undefined behavior. The C 1999 standard says, in 6.7.8 10, “If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.” (This paragraph goes on to define how static objects are initialized, so the only uninitialized objects we are concerned about are automatic objects.)

3.17.2 defines “indeterminate value” as “either an unspecified value or a trap representation”. 3.17.3 defines “unspecified value” as “valid value of the relevant type where this International Standard imposes no requirements on which value is chosen in any instance”.

So, if the uninitialized unsigned int x has an unspecified value, then x -= x must produce zero. That leaves the question of whether it may be a trap representation. Accessing a trap value does cause undefined behavior, per 6.2.6.1 5.

Some types of objects may have trap representations, such as the signaling NaNs of floating-point numbers. But unsigned integers are special. Per 6.2.6.2, each of the N value bits of an unsigned int represents a power of 2, and each combination of the value bits represents one of the values from 0 to 2^N-1. So unsigned integers can have trap representations only due to some values in their padding bits (such as a parity bit).

If, on your target platform, an unsigned int has no padding bits, then an uninitialized unsigned int cannot have a trap representation, and using its value cannot cause undefined behavior.