Bit shift leads to strange type conversion

Question

The following code compiles without warning:

std::uint16_t a = 12;
std::uint16_t b = a & 0x003f;

However, performing a bit shift along with the bitwise and leads to an 'implicit cast warning':

std::uint16_t b = (a & 0x003f) << 10; // Warning generated.

Both gcc and clang complain that there is an implicit conversion from int to uint16_t, yet I fail to see why introducing the bit shift would cause the right hand expression to suddenly evaluate to an int.

EDIT: For clang, I compiled with the -std=c++14 -Weverything flags; for gcc, I compiled with the -std=c++14 -Wall -Wconversion flags.

Answer 1

Doing any arithmetic with integer types always is preceded by promotion to at least (sometimes, but not in this case using gcc, unsigned) int. As you can see by this example, this applies to you first, warning-less variant too.

The probably best way to get around these (admittedly often surprising) integer promotion rules would be to use an unsigned int (or uint32_t on common platforms) from the get go.

If you cannot or don't want to use the bigger type, you can just static_cast the result of the whole expression back to std::uint16_t:

std::uint16_t b = static_cast<std::uint16_t>((a & 0x003f) << 10); 

This will correctly result in the RHS-value mod 2^16.