On Wed, Aug 21, 2019 at 6:38 PM sdkrystian via Std-Proposals <std-proposals@lists.isocpp.org> wrote:
> It's nothing to do with the value of the int object as a whole, but of the leftmost byte. Why on earth would it not depend on endianness?

Because the standard says so. Here are some citations:

In that particular cast, the value of the pointer is unchanged. http://eel.is/c++draft/expr.static.cast#13.sentence-4

Since the value is unchanged, it still points to the int object. Therefore, when the indirection operator is used, it yields an lvalue of type `char` that denotes the object the pointer points to. http://eel.is/c++draft/expr.unary.op#1.sentence-1

This part is fine.

Now, when an lvalue-to-rvalue conversion is applied to the lvalue (such as during initialization), the result is the value of the object (which has NOT been truncated), and since that value is within the representable range of `char`, the behavior is well defined, and will ALWAYS yield 42. 

This part does not follow.

(If it's outside the representable range of char, the behavior is undefined http://eel.is/c++draft/expr.pre#4)

On a 32-bit little endian system, the value 42 is represented in memory as [42, 0, 0, 0].
On a 32-bit big endian system, the value 42 is represented in memory as [0, 0, 0, 42].

The char* that you're getting is pointing to the first char, which can have either value. As the text you're citing states, the value of the char* must be the value of the int* - so it must point to the first byte in memory. Which byte is in that slot depends on the endianness of the system.

In order for the char* dereference to give you 42 on a big endian system, the cast we do would've had to move the pointer value. But the pointer value must remain unchanged.