64位机器中的最大文件映射大小是多少 [英] What's the max file mapping size in 64bits machine

问题描述

我是 64 位架构的新手.你能告诉我 64 位 linux 机器中文件映射支持的最大文件大小是多少.我想通过文件映射打开20GB以上的文件，可以吗?

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?

我写了一个示例代码.但是当我在 GBSIZE 偏移量中获取指针的值时，它会导致 Bus Error:

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); 

顺便说一句，printf("%c",*pBegin); 工作正常.和我的地址大小:38 位物理，48 位虚拟

BTW, printf("%c",*pBegin ); works fine. and my address sizes : 38 bits physical, 48 bits virtual

完整代码如下:

#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;
}

推荐答案

尽管指针是 64 位宽，但大多数处理器实际上并不支持使用完整 64 位的虚拟地址.要查看您的处理器支持的虚拟地址大小，请查看 /proc/cpuinfo(通常为 48 位).

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

此外，一半的虚拟地址空间由内核使用，用户空间不可用 - 在当前的 Linux 实现中留下 47 位.

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.

然而，即使考虑到这一点，您仍然有足够空间来存放 20GB 的文件.理论上47位意味着128TB的虚拟地址空间.

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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