我希望直接从我的C代码中获取大页面大小的值,而无需运行bash命令。
从bash我可以做到这一点
grep pse /proc/cpuinfo > /dev/null && echo '2M huge page size are supported'
grep pdpe1gb /proc/cpuinfo> /dev/null && echo '1G huge page size are supported'
其次如何使用1G巨大页面大小的mmap?
感谢
更新
摘录代码
#include <stdio.h>
#include <limits.h>
#include <hugetlbfs.h>
int main(void){
long result1 = gethugepagesize();
printf( "%d\n", result1 );
long result2 = gethugepagesizes( NULL, 0);
printf( "%d\n", result2 );
long result3 = getpagesizes( NULL, 0);
printf( "%d\n", result3 );
printf("%d\n", PF_LINUX_HUGETLB);
return 0;
}
输出
2097152
1
2
1048576
这里gethugepagesize返回2 Mb 1Gb大页面怎么样?
答案 0 :(得分:3)
试试这个。
#include <hugetlbfs.h>
int getpagesizes(long pagesizes[], int n_elem);
答案 1 :(得分:1)
由于我没有足够的声誉点发表评论,我会在答案中发布。
我修改了你的代码:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <hugetlbfs.h>
void dump_page_sizes_arr( int (getter_fn)(long pagesizes[], int n_elem), int elem ) {
int elem_alloc, i;
long *pagesizes = NULL;
if( elem <= 0 ) return;
pagesizes = calloc( elem, sizeof(long) );
if( pagesizes == NULL ) return;
elem_alloc = getter_fn( pagesizes, elem );
if ( elem_alloc != elem ) goto stop;
for( i=0; i<elem_alloc; i++ ) printf( " %ld\n", pagesizes[i] );
stop:
free( pagesizes );
}
int main(void){
long result1 = gethugepagesize();
printf( "huge page size = %ld\n", result1 );
int result2 = gethugepagesizes( NULL, 0 );
printf( "huge page sizes [%d] =\n", result2 );
dump_page_sizes_arr( gethugepagesizes, result2 );
int result3 = getpagesizes( NULL, 0 );
printf( "page sizes [%d] =\n", result3 );
dump_page_sizes_arr( getpagesizes, result3 );
printf( "PF_LINUX_HUGETLB = %d\n", PF_LINUX_HUGETLB );
return 0;
}
我的系统似乎得到了非常相似的结果(Mint 17.3 x64 + 2 GiB RAM):
huge page size = 2097152
huge page sizes [1] =
2097152
page sizes [2] =
4096
2097152
PF_LINUX_HUGETLB = 1048576
所以回答你的问题:
1Gb大页面怎么样?
......您的系统似乎不支持它。