Question

我有一个C程序，用C计算页面错误服务时间。对于这个程序，我有2个大文件（每个小于3GB - 几乎与RAM的大小）

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <sys/stat.h>
#include <fcntl.h>
#include "rdstc.h"
#include "config.h"

#define KB 1024
#define MB 1024 * KB
#define GB 1024 * MB
#define SIZE_OF_MEMORY 1 * GB   // Main memory size

#define handle_error(msg) do { perror(msg); exit(EXIT_FAILURE); } while (0)

int main(int argc, char *argv[]){

    int fd1, fd2;
    char *addr1, *addr2, c;
    int i, j;
    long long unsigned int s_t, e_t, t=0;

    if (argc != 3){
        printf("usage: a.out <file1> <file2> \n");
        exit(EXIT_FAILURE);
    }

    if ((fd1 = open(argv[1], O_RDONLY)) == -1){
        handle_error("open");
    }

    if ((fd2 = open(argv[2], O_RDONLY)) == -1){
        handle_error("open");
    }

    posix_fadvise(fd1, 0, 0, POSIX_FADV_RANDOM);
    posix_fadvise(fd2, 0, 0, POSIX_FADV_RANDOM);

    addr1 = (char *) mmap(0, SIZE_OF_MEMORY, PROT_READ, MAP_PRIVATE | MAP_POPULATE, fd1, 0);
    if (addr1 == MAP_FAILED){
        handle_error("mmap");
    }
    addr2 = (char *) mmap(0, SIZE_OF_MEMORY, PROT_READ, MAP_PRIVATE | MAP_POPULATE, fd2, 0);
    if (addr2 == MAP_FAILED){
        handle_error("mmap");
    }

    madvise(addr1, 0, MADV_RANDOM);
    madvise(addr2, 0, MADV_RANDOM);

    j = 32;     // default read ahead size if 256 blocks (assuming each block is of 512 bytes)
    for(i = 0; i < ITERATIONS; i++){
        s_t = rdtsc();
            c = addr1[i + j*4*KB];      // read at multiple of page size, so every read causes a page fault
            j *= 2;
        e_t = rdtsc();
        t += (e_t - s_t);
    }
    printf("Time required to service a page faut is %f \n", (t/ITERATIONS)/CPU_FREQ);

    munmap(addr1, SIZE_OF_MEMORY);
    munmap(addr2, SIZE_OF_MEMORY);

    return 0;
}

我收到以下编译器警告：

lmelvix@Melvix:~/projects/mem$ gcc mem1_4.c -lm
mem1_4.c: In function ‘main’:
mem1_4.c:11:17: warning: integer overflow in expression [-Woverflow]
 #define MB 1024 * KB
                ^
mem1_4.c:12:19: note: in expansion of macro ‘MB’
 #define GB 1024 * MB
                ^
mem1_4.c:13:28: note: in expansion of macro ‘GB’
 #define SIZE_OF_MEMORY 2 * GB   // Main memory size
                            ^
mem1_4.c:40:30: note: in expansion of macro ‘SIZE_OF_MEMORY’
    addr1 = (char *) mmap(0, SIZE_OF_MEMORY, PROT_READ, MAP_PRIVATE | MAP_POPULATE, fd1, 0);
                            ^
mem1_4.c:11:17: warning: integer overflow in expression [-Woverflow]
 #define MB 1024 * KB
                ^
mem1_4.c:12:19: note: in expansion of macro ‘MB’
 #define GB 1024 * MB
                ^
mem1_4.c:13:28: note: in expansion of macro ‘GB’
 #define SIZE_OF_MEMORY 2 * GB   // Main memory size
                            ^
mem1_4.c:44:30: note: in expansion of macro ‘SIZE_OF_MEMORY’
    addr2 = (char *) mmap(0, SIZE_OF_MEMORY, PROT_READ, MAP_PRIVATE | MAP_POPULATE, fd2, 0);
                            ^
mem1_4.c:11:17: warning: integer overflow in expression [-Woverflow]
 #define MB 1024 * KB
                ^
mem1_4.c:12:19: note: in expansion of macro ‘MB’
 #define GB 1024 * MB
                ^
mem1_4.c:13:28: note: in expansion of macro ‘GB’
 #define SIZE_OF_MEMORY 2 * GB   // Main memory size
                            ^
mem1_4.c:62:19: note: in expansion of macro ‘SIZE_OF_MEMORY’
    munmap(addr1, SIZE_OF_MEMORY);
                ^
mem1_4.c:11:17: warning: integer overflow in expression [-Woverflow]
 #define MB 1024 * KB
                ^
mem1_4.c:12:19: note: in expansion of macro ‘MB’
 #define GB 1024 * MB
                ^
mem1_4.c:13:28: note: in expansion of macro ‘GB’
 #define SIZE_OF_MEMORY 2 * GB   // Main memory size
                            ^
mem1_4.c:63:19: note: in expansion of macro ‘SIZE_OF_MEMORY’
    munmap(addr2, SIZE_OF_MEMORY);
                ^

当我使用命令运行它时，我收到错误

./a.out file1.txt file2.txt
mmap: Cannot allocate memory

代码是做什么的？我们使用标志

映射这两个文件

MAP_PRIVATE（以便其他进程不应访问此文件）和MAP_POPULATE（以便

当我们调用mmap（）完整文件映射到内存中）以及PROT_READ保护标志。

首先我们映射file1，因为我们使用MAP_POPULATE完整RAM由与该文件对应的数据填充。在此之后，我们使用相同的标志映射file2，因此现在我们将file2完全映射到RAM中。因此，访问file1的数据将导致页面错误，因为file2占用了所有可用的RAM。我们还调用带有MADV_RANDOM标志的madvise（）系统调用，建议内核不要对这两个文件进行页面预读。所以现在一旦这个初始设置完成，file2占用了所有可用的RAM，我们随机访问对应于file1的数据（以避免由内核执行的预读读取优化的任何影响，并且还避免从L3缓存读取）因为，RAM充满了对应于file2的数据，每次访问对应文件的数据都会导致页面错误。我们在循环中的映射区域上执行10个随机读数，并测量此操作所需的平均时间。

Answer 1

看看你得到的编译器警告。你在这里有一个整数溢出：#define SIZE_OF_MEMORY 2 * GB。这等于2^31 == 0b1000 ... 0，对于signed int等于INT_MIN。这就是mmap失败的原因。

您应该在定义中使用unsigned literals：

#define KB (1024u)
#define MB (1024u * KB)
#define GB (1024u * MB)
#define SIZE_OF_MEMORY (2u * GB)

mmap：无法分配内存

1 个答案: