什么是最大文件大小映射在64位机 [英] What's the max file mapping size in 64bits machine
问题描述
我是新来的64位架构。你能告诉我什么是按文件映射在64位Linux机器支持的最大文件大小。我想通过文件映射打开超过20GB的文件,是不是可以?
我写了一个样本code。但它会导致总线错误当我在GBSIZE指针的偏移值:
无符号字符* pCur = pBegin + GBSIZE;
// pBegin是MMAP返回的指针
的printf(%C,* pCur);
顺便说一句,的printf(%C,* pBegin);
工作正常。和我的地址尺寸:38位物理,48位虚拟
下面是完整的code:
的#include< stdio.h中>
#包括< stdlib.h中>
#包括< SYS / types.h中>
#包括< SYS / stat.h>
#包括< unistd.h中>
#包括< fcntl.h>
#包括< SYS / mman.h>
//#定义FILEPATH小文件
#定义FILEPATH大文件
#定义GBSIZE(1024L * 1024L * 1024L)
#定义TBSIZE(1024L * GBSIZE)
#定义NUMSIZE(20L * GBSIZE)
//#定义NUMSIZE(10)
#定义FILESIZE(NUMINTS *的sizeof(INT))
INT主(INT ARGC,字符* argv的[])
{
INT I;
INT的fd;
无符号字符* pBegin;
FD =打开(FILEPATH,O_RDONLY);
如果(FD == -1){
PERROR(错误打开文件进行读取);
出口(EXIT_FAILURE);
}
pBegin = MMAP(0,NUMSIZE,PROT_READ,MAP_SHARED,FD,0);
如果(pBegin == MAP_FAILED){
关闭(FD);
PERROR(错误mmapping文件);
出口(EXIT_FAILURE);
}
/ **错误发生在这里! ** /
无符号字符* pCur = pBegin + GBSIZE;
的printf(%C,* pCur);
如果(在munmap(pBegin,NUMSIZE)== -1){
PERROR(错误未mmapping文件);
}
关闭(FD);
返回0;
}
虽然指针是64位宽,大多数处理器实际上并不支持使用完整的64位虚拟地址。要查看的/ proc / cpuinfo的
什么尺寸的虚拟地址,你的处理器支持,看看(48位是典型的)。
grep的地址尺寸的/ proc / cpuinfo的
此外,一半的虚拟地址空间中所使用的内核,而不是提供给用户空间 - 留在目前的Linux实现47位
。不过,即使考虑到这一点,你将仍然有大量的空间为20GB的文件。 47位在理论上意味着128TB的一个虚拟地址空间。
I'm new to 64-bits architecture. Could you tell me what's MAX file size supported by file mapping in 64 bits linux machine. I want to open more than 20GB files by file mapping, is it available?
I write a sample code. But it causes Bus Error when I get the value of the pointer in GBSIZE offset:
unsigned char* pCur = pBegin + GBSIZE;
//pBegin is the pointer returned by mmap
printf("%c",*pCur);
BTW, printf("%c",*pBegin );
works fine. and my address sizes : 38 bits physical, 48 bits virtual
Here is the full code:
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <fcntl.h>
#include <sys/mman.h>
//#define FILEPATH "smallfile"
#define FILEPATH "bigfile"
#define GBSIZE (1024L*1024L*1024L)
#define TBSIZE (1024L*GBSIZE)
#define NUMSIZE (20L * GBSIZE)
//#define NUMSIZE (10)
#define FILESIZE (NUMINTS * sizeof(int))
int main(int argc, char *argv[])
{
int i;
int fd;
unsigned char *pBegin;
fd = open(FILEPATH, O_RDONLY);
if (fd == -1) {
perror("Error opening file for reading");
exit(EXIT_FAILURE);
}
pBegin = mmap(0, NUMSIZE, PROT_READ, MAP_SHARED, fd, 0);
if (pBegin == MAP_FAILED) {
close(fd);
perror("Error mmapping the file");
exit(EXIT_FAILURE);
}
/** ERROR happens here!!! **/
unsigned char* pCur = pBegin + GBSIZE;
printf("%c",*pCur);
if (munmap(pBegin, NUMSIZE) == -1) {
perror("Error un-mmapping the file");
}
close(fd);
return 0;
}
Although pointers are 64-bit wide, most processors do not actually support virtual addresses using the full 64 bits. To see what size virtual addresses your processor supports, look in /proc/cpuinfo
(48 bits is typical).
grep "address sizes" /proc/cpuinfo
Additionally, half of the virtual address space is used by the kernel and not available to userspace - leaving 47 bits in the current Linux implementation.
However, even taking this into account, you will still have plenty of room for a 20GB file. 47 bits in theory means a virtual address space of 128TB.
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