Why no warning when uint32 casts to uint8 with gcc C

c, gcc, gcc-warning, implicit-conversion, type-conversion

Solution

There is no warning because the expression:

((uint32_val >> 8) & 0x000000FF)

is always convertible to a type that fits in a byte. Your uint32_val is not cast, but takes part in a calculation. The compiler can optimize this, and in this case it is clear the result after bit-wise and-ing with `000000ff` will always fit the provided l-value type, hence no warning.

Problem

Compiling with -Wconversion, ``` uint32_val = 0x00000C00; uint_8_val = ((uint32_val >> 8) & 0x000000FF); ``` gives no warning. What compiler flag shall I use to get a warning?

Original source