Hi, I'm using a 2.6.26 kernel on an AMD64 machine. Im running some timing experiments on the Linux swapping system. My system has 2GB of RAM but I boot linux with mem=1GB. There's is some behavior happening which I don't understand and I'm hoping someone can explain it to me. Here is what my experimental (simple) programs do. in sequence : 1) Process A grabs a lot of memory (calloc). It takes about 90% of memory 2) Process A does some reads/writes to all its allocated memory 3) Start process B. This also grabs a lot of memory (calloc). 4) Process B writes to all its allocated memory 5) OS swaps out Process A pages (to swap device) to make room for B 6) Process A does a timed read of all its pages. In this case I was expecting it to take a while to do the timed read since all its pages have been swapped to disk. However this is not the case. It takes the same amount of time as when I don't run Process B. This is counter to my understanding of how it should work. If I replace step 6) above with : 6) Process A does a timed write of all its pages Then I see the expected behavior which is that it takes a long time because all its pages are swapped in first. Why is there this discrepancy between timed reads and writes ? At first I thought it may be because AMD64 has a tagged cache and tagged TLB which causes this behaviour but I'm not really sure. Does anyone have an idea ? Thanks arn -- To unsubscribe from this list: send an email with "unsubscribe kernelnewbies" to ecartis@xxxxxxxxxxxx Please read the FAQ at http://kernelnewbies.org/FAQ
