Anatomy of a Program in Memory (2009)

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After reading many articles about virtual memory and how kernel space is mapped into every process I don&#x27;t understand why it is necessary. Why can&#x27;t process only have mapped it&#x27;s user mode space? Also it only seems to be case of unixes and windows. Not sure how exactly it&#x27;s done in OSX but &quot;Mac OS X does not map the kernel into each user address space, and therefore each user&#x2F;kernel transition (in either direction) requires an address space switch.&quot; <a href="https:&#x2F;&#x2F;flylib.com&#x2F;books&#x2F;en&#x2F;3.126.1.91&#x2F;1&#x2F;" rel="nofollow">https:&#x2F;&#x2F;flylib.com&#x2F;books&#x2F;en&#x2F;3.126.1.91&#x2F;1&#x2F;</a>

The author states:<p>&gt;It is also possible to create an anonymous memory mapping that does not correspond to any files, being used instead for program data.&quot;<p>This isn&#x27;t strictly true though is it? It was my understanding even mmap() MAP_ANONYMOUS used a file interface, and that the way the kernel creates anonymous maps is by creating an instance of &#x2F;dev&#x2F;zero in tmpfs. Although I believe the file descriptor might be ignored however.

if the process depicted in the diagram were to start a second thread, where would that second thead’s stack go in the diagram? The two threads would share the same heap.

&gt;&gt; In Linux, kernel space is constantly present and maps the same physical memory in all processes.<p>That&#x27;s right there together with the city states of Greece and other ancient memories. Meltdown and Specter happened.

The title needs 2009 in it, since this article is from that year.