Why does my C++ code with a type mismatch compile without warnings using the GCC "-Wall" flag?

I have the following code:

#include <iostream>

char add(char a, int b)
{
    return a + b;
}

int main(void)
{
    long long a = 100000000;
    long long b = 20;

    int x = add(a, b);

    std::cout << x << std::endl;

    return 0;
}

And I don't understand why does my code compile without any warnings.

Shouldn't the GCC compiler with the -Wall flag give me a type mismatch error?

This is not an error it is a valid implicit conversion, although it involves narrowing the type. You can get warning about this with e.g. -Weverything in clang which will emit

<source>:13:17: warning: implicit conversion loses integer precision: 'long long' to 'char' [-Wimplicit-int-conversion]
   13 |     int x = add(a, b);
      |             ~~~ ^
<source>:13:20: warning: implicit conversion loses integer precision: 'long long' to 'int' [-Wshorten-64-to-32]
   13 |     int x = add(a, b);
      |             ~~~    ^