the answer you are looking for to your core question is “No”. Generally all kernels will assume _some_ backing store. The reason for this is dealing with RAM b

the answer you are looking for to your core question is “No”. Generally all kernels will assume _some_ backing store. The reason for this is dealing with RAM being full. This is because updating MMU is generally expensive, so it’s easier to mark an address as belonging to your process, and commit the pages after the fact. Now this can be changed in settings. Ensure pages are backed before allocated, and running swap off can be done. But then EVERY and let me repeat EVERY SINGLE piece of code you run has to able to handle malloc failing gracefully. And next to none do. This is a strange condition because how you handle it depends on your kernel configuration, and how much stack you have left. Which is difficult to know unless you wrote every piece of code FOR your custom kernel config. Ultimately just having swap enabled and assuming a back store is easier. The few places where you get away with 100% ram and a read only store is embedded (generally, not always) and in these conditions you can closely tie software to kernel configuration.