问题
I am reading about logical and physical addressing. I am confused that when a binary file is run, does it pass first through the CPU where the logical address are generated or is it directly copied to the physical memory ?
回答1:
when a binary file is run, does it pass first through the CPU where the logical address are generated or is it directly copied to the physical memory ?
Typically some code somewhere loads the executable file's headers into memory, and then uses information from the headers to figure out where various pieces of the file (sections - e.g. .text, .data, etc) should be in virtual memory and what each virtual page's virtual permissions should be (if writes are allowed, if execution is allowed).
After this, areas of the virtual address space are set up. Often this is done by memory mapping the relevant part of the file into the virtual address space, without actually loading them into physical memory. In this case each page's actual permissions don't reflect the page's virtual permissions (e.g. a "read/write" page might be "not present" initially, and when software tries to read from the page you'll get a page fault and the page fault handler might fetch the page from disk and change the page to "present, read only"; and then later when software tries to write to the page you might get a second page fault and the page fault handler might do a "copy on write" so that anything else using the same physical page isn't effected and then make the new copy "read/write" so that it matches the original virtual permissions).
While this is happening; the OS could (depending on amount of free physical RAM and whether storage devices have more important data to transfer) be prefetching the file data from disk (e.g. into VFS cache), and could be "opportunistically" updating the process' page tables to avoid the overhead of page faults for pages that have been prefetched.
However; if the OS knows that the file was on unreliable and/or removable media it may decide that using memory mapped files is a bad idea and may actually load the needed executable's sections into memory before executing it; and an OS could have other features that cause the file to be loaded into RAM before it's executed (e.g. if the OS checks that an executable file's digital signature is correct before allowing the file to be executed, then the entire file probably needs to be loaded into memory to allow the digital signature can be checked, and in that case the entire file is likely to still be in memory when virtual address space is set up).
回答2:
You need to read an entire book on these topics and spend several weeks on that
But Operating Systems: Three Easy Pieces is a good book, and it is freely downloadable.
Once you have read it, look perhaps also into osdev.org for practical things. And don't forget free software OSes such as Linux, e.g. https://kernelnewbies.org/
Be aware of copy-on-write and virtual address space...
回答3:
An executable files is generally an interpreted program itself that is executed by the loader. The executable contains instructions that tell the loader how the program should exist in VIRTUAL memory. By that, I mean the instructions in the executable define how the initial VIRTUAL representation of process address space.
So when the executable starts, there is only a virtual representation of the address space in secondary storage. As the program executes, it starts page faulting repeatedly to load pages into memory. After the initial load, the page fault rate dies down.
The executable NORMALLY only contains logical addresses.
来源：https://stackoverflow.com/questions/55786222/how-is-a-program-loaded-in-os