Implicit type conversion for bitwise operation

Question

I have this operation:

uint32_t DIM = // ...
int32_t x = // ...

// Operation:
x & (DIM-1u)

How does implicit type conversion work in the statement x & (DIM-1u)?

Rules can be found here: https://en.cppreference.com/w/c/language/conversion#Usual_arithmetic_conversions — NathanOliver, Jul 09 '20 at 15:35
related/dupe: https://stackoverflow.com/questions/46073295/implicit-type-promotion-rules — NathanOliver, Jul 09 '20 at 15:35

Bathsheba · Accepted Answer · 2020-07-09T15:51:18.367

Two scenarios, noting that 1u is a literal of type unsigned:

unsigned is in the inclusive range of 16 bits to 31 bits. The type of DIM - 1u is uint32_t, and the whole expression is uint32_t. This is because the signed type in a binary expression where the other argument is an unsigned type is converted implicitly to unsigned.
unsigned is 32 bits or larger. Then the type of DIM - 1u is unsigned, and the same for the type of the whole expression.

Finally, note that the C++ standard permits unsigned and std::uint32_t to be the same type; i.e.

std::cout << std::is_same<std::uint32_t, unsigned>::value;

is allowed to be 1.

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