Comment by tialaramex

It won't "overflow memory".

This says there will be an immutable array of six bytes, with the ASCII letters for "thing" in the first five and then the sixth is zero, this array can be coerced to the pointer type char* (a pointer to bytes) and then (though a modern C compiler will tell you this is a terrible idea) coerced to the signed integer type int.

The six byte array will end up in the "read only data" section of the executable, it doesn't "overflow memory" and isn't stored in the x. Even if you gave x a more sensible type "char*" that word "thing" isn't somehow stored in your variable, it's a pointer.

So, this isn't the same at all and you don't understand C as well as you thought you did.

Edited: fix escaping bold markers