There are two types of pages: anonymous and mapped from files. The code pages are mapped from a binary; they are not very special.

Under memory pressure, the kernel evicts less popular pages from memory. If a page has been mapped from a file, it is dropped (if dirty, then it is written out first). If it is needed later, the kernel can read it back from the file. If a page is anonymous (read: heap page), then there is no backing file and the kernel copies it to swap before dropping it. This is swapping.

So, what happens if you disable swap and the kernel is low on memory? What can it evict? Anonymous pages cannot be evicted: there is no swap to put a copy in. The only choice the kernel has is to evict pages that are mapped from files. Those include pages mapped from the executable. You don't eliminate stalls by disabling swap, you just move them elsewhere: the kernel will page out code and your app gets paused whenever the execution flow hits such a page.

Thanks for the clarification, learning everyday.

I haven't had to analyze the performance of no-swap processes before. My assumption is that code is hot enough to avoid eviction and that evicted code pages are rather the exception. To strong-man the argument, I can imagine long running complex (bloated) services could have parts that are not touched unless a specific request comes in.

>So, what happens if you disable swap and the kernel is low on memory? What can it evict?

Your userspace early OOM killer triggers and lets you know you're trying to run more than will fit in memory so you don't do it again. (In my experience, the kernel OOM killer can't be trusted to kill processes soon enough.)