Why the extra leading zeroes? If you want to ensure the literal is treated like a U32, you can do 0x00000001 or simply 1U, but I suspect the compiler will default to 32 bits anyway.
Why the extra leading zeroes? If you want to ensure the literal is treated like a U32, you can do 0x00000001 or simply 1U, but I suspect the compiler will default to 32 bits anyway.