memory-management 相关问题

static int __init hello_entry(void){ unsigned long i = 0; unsigned long j = 0; unsigned long counter; int fault = 0; char default_value = 0xff; unsigned long size_of_memory_to_test_bytes = 10 * 1024 * 1024;//10MB chunk unsigned long page_size = 8 * 1024;//8kB page size. From data sheet for elpida 4GB ddr3 ram unsigned long number_of_pages = size_of_memory_to_test_bytes / page_size; unsigned long number_of_hammer_access = 200000; printk("Number Of Bytes to allocate: %ld\n", number_of_pages * page_size); char* pages = (char *)alloc_pages_exact(number_of_pages * page_size, GFP_KERNEL); while(pages == 0);//The Program must wait till enough pages have been allocated! printk("Starting Address: %x", pages); char *hammerRow0, *hammerRow1, *victimRow; char hammerRow0_data, victimRow_data, hammerRow1_data, read_midvalue; printk(KERN_ALERT "Memory Test Entry!\n"); printk(KERN_ALERT "Page Size: %ld\n", page_size); printk(KERN_ALERT "Cache Line Size: %d\n", cache_line_size()); //The design consists of two hammer rows which will be repeatedly accessed one after another //There will be a victim row in between the two hammer rows, kind of like a victim row sandwiched between two hammer rows //This will server two purposes: //1. Greater probability of inducing bitflip in the victim row due to repeated access of two adjacent rows. //2. Clearing out the row buffer of DRAM chip so that it is guaranteed that the wordline will be accessed to access the data printk(KERN_INFO "Rowhammer Test Starts\n"); for(counter = 0; counter < number_of_pages - 2; counter++){ //Doing the rowhammer test on every page of dram hammerRow0 = pages + counter; victimRow = hammerRow0 + page_size; hammerRow1 = victimRow + page_size; //Storing initial data into rows to check them later for errors for(i = 0; i < page_size / sizeof(char); i++){ //Setting all bits of dram 1 to maximize the possibility of charge leak hammerRow0[i] = default_value; victimRow[i] = default_value; hammerRow1[i] = default_value; } //Starting Hammer Test //printk(KERN_INFO "Rowhammer Test Starts\n"); fault = 0; for(j = 0; j < number_of_hammer_access; j++){ //hammerRow0 access //This is an x86 specific funnction that executes CLFLUSH. Mre info on CLFLUSh: https://c9x.me/x86/html/file_module_x86_id_30.html clflush_cache_range(hammerRow0, sizeof(char)); read_midvalue = hammerRow0[0]; //Reading the first value present in the page //hammerRow1 access clflush_cache_range(hammerRow1, sizeof(int)); read_midvalue = hammerRow1[0]; //Reading the first value present in the page } //Checking the victim row for errors for(i = 0; i < page_size / sizeof(int); i++){ victimRow_data = victimRow[i]; if(victimRow_data != default_value){ printk(KERN_ALERT "victimRow Fault (After Hammering) at index %d: Present Value: %x \t Expected Value: %x:\tRow numer %d\n", i, victimRow_data, 0xff, counter); fault = 1; } } if(fault == 1){ printk(KERN_ALERT "Rowhammer Test Failed\n"); return 0; } else { //printk(KERN_ALERT "Rowhammer Test Passed for Row %d\n", counter); } } printk(KERN_ALERT "Rowhammer Test Passed\n"); return 0; } static void __exit hello_exit(void){ printk(KERN_INFO "Memory Test Exit!\n"); } module_init(hello_entry); module_exit(hello_exit); //Licensing Info MODULE_LICENSE("GPL"); MODULE_AUTHOR("jonvonton"); MODULE_DESCRIPTION("Rowhammer Testing");

c linux memory-management linux-kernel kernel-module

回答 1 投票 0

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如何释放被占用但尽快占据大物体的内存？

c# .net memory-management garbage-collection clr

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为什么在静态内存部分中分配堆栈帧，如果他们没有打电话？为什么表现会增长？

对我的问题进行搜索只会产生无关的结果。这是最接近的，但没有回答我的问题是堆栈“静态分配”的变量？所以我知道如果是

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