[kccqzy]

Well there's the so-called usual arithmetic conversions that will basically convert every uint16_t to an unsigned int. The C and C++ languages do a silent conversion on your back anyways so you might as well make it explicit.

[loeg]

Usually promotions are to signed int, not unsigned. (With some exceptions. But every uint16_t value can fit in int.)

[Filligree]

Unless int is 16-bit. Code like this is potentially UB; you should use int32_t as the target.

[kevin_thibedeau]

You should use long, and don't ever assume it's exactly 32-bits. The fixed size types are often an overused crutch that hampers future portability.

[loeg]

There are no mainstream 16-bit int platforms. It's fine to know what you target.

The promotions that are really surprising are from uint64_t bitfields to int (because it's based on value representability).

  struct {
    uint64_t a : 33,
             b : 15;
  } s;
  // s.b gets "promoted" to int, s.a does not

[inetknght]

A well-configured C++ compiler will error-out on such a silent conversion.

[kccqzy]

The C++ compiler is required to perform this silent conversion according to the standard: https://en.cppreference.com/w/cpp/language/implicit_conversi...