Question

我无法在具有足够内存的64位计算机中使用mmap分配超过2GB的连续内存。以下代码愉快地运行。但是，当我将机器中连续mmap的页面数量增加到略高于2GB时，mmap会失败。

#include <sys/mman.h>
#include <stdio.h>
#include <signal.h>
#include <stdlib.h>
#include <time.h>
#include <errno.h>
#include <unistd.h>

#define PAGE_SIZE (4*1024)
#define START_ADDRESS 0x8000000
#define START_ADDRESS2 0x800000000
#define N_ITERATIONS 524288

// one more iteration and mmap fails
//#define N_ITERATIONS 524289

void allocate(void* base_address)
{
    int i;
    for (i = 0; i < N_ITERATIONS; ++i) {
        void* current_addr = base_address + PAGE_SIZE * i;
        void* ret = mmap((void*)current_addr, PAGE_SIZE, PROT_EXEC|PROT_READ|PROT_WRITE, 
                MAP_NORESERVE | MAP_FIXED | MAP_PRIVATE|MAP_ANONYMOUS, 0, 0); 
        if (ret == MAP_FAILED) {
            fprintf(stderr, "Error mmap. errno: %d\n", errno);
            exit(-1);
        }   
        printf("%d\n", i); 
    }   
}

int main()
{
    allocate((void*)START_ADDRESS);
    allocate((void*)START_ADDRESS2);
    return 0;
}

ulimit -a

-t: cpu time (seconds)              unlimited
-f: file size (blocks)              unlimited
-d: data seg size (kbytes)          unlimited
-s: stack size (kbytes)             8192
-c: core file size (blocks)         unlimited
-m: resident set size (kbytes)      unlimited
-u: processes                       515550
-n: file descriptors                65535
-l: locked-in-memory size (kbytes)  unlimited
-v: address space (kbytes)          unlimited
-x: file locks                      unlimited
-i: pending signals                 515550
-q: bytes in POSIX msg queues       819200
-e: max nice                        0
-r: max rt priority                 0
-N 15:                              unlimited

什么阻止内核分配超过2GB的连续数据？

Answer 1

base_address PAGE_SIZE * i 0xffffffff86000000。{/ p>

因此，您试图在i附近的地址上分配。

为fscanf代替%s，它会起作用。

Mmap无法分配超过2GB的连续数据

1 个答案: