Unexpected (IMO) constant-conversion warning

consider following code:

#include <iostream>

static constexpr uint8_t a = 0x80;
static constexpr uint8_t b = ~a;

It produces the following warning:

<source>:5:30: warning: implicit conversion from 'int' to 'const uint8_t' (aka 'const unsigned char') changes value from -129 to 127 [-Wconstant-conversion]

static constexpr uint8_t b = ~a;

I do not understand, why overflow warning is here, there is no explicite int in this code. I am operating on variables with the same type. When removing constexpr, warning disappears, when changing to:

static constexpr uint8_t b = uint8_t{~a}; 

warning disappears. So looks like ~ operator implicite changes variable to int?

Could someone explain it to me?