Unsigned values in C

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Assign a int -1 to an unsigned: As -1 does not fit in the range [0...UINT_MAX], multiples of UINT_MAX+1 are added until the answer is in range. Evidently UINT_MAX is pow(2,32)-1 or 429496725 on OP’s machine so a has the value of 4294967295.

unsigned int a = -1;

The "%x", "%u" specifier expects a matching unsigned. Since these do not match, “If a conversion specification is invalid, the behavior is undefined.
If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.” C11 ยง7.21.6.1 9. The printf specifier does not change b.

printf("%x\n", b);  // UB
printf("%u\n", b);  // UB

The "%d" specifier expects a matching int. Since these do not match, more UB.

printf("%d\n", a);  // UB

Given undefined behavior, the conclusions are not supported.

both cases, the bytes are the same (ffffffff).

Even with the same bit pattern, different types may have different values. ffffffff as an unsigned has the value of 4294967295. As an int, depending signed integer encoding, it has the value of -1, -2147483647 or TBD. As a float it may be a NAN.

what is unsigned word for?

unsigned stores a whole number in the range [0 ... UINT_MAX]. It never has a negative value. If code needs a non-negative number, use unsigned. If code needs a counting number that may be +, – or 0, use int.

Update: to avoid a compiler warning about assigning a signed int to unsigned, use the below. This is an unsigned 1u being negated – which is well defined as above. The effect is the same as a -1, but conveys to the compiler direct intentions.

unsigned int a = -1u;

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