为什么没有malloc的填充内存？ | 为什么没有malloc的填充内存

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问题描述

我有以下的code：

#include <iostream>
#include <stdlib.h>
#include <stdio.h>

int main() {
    int data = 0;
    char *byte = (char *)malloc(sizeof(char)*1000000000);
byte[999999999] = 'a';
printf("%c",byte[999999999]);
    scanf("%d",&data);
    return 0;
}

在内存中寻找程序启动和scanf函数之前，我会期望内存增加1 GB之前。为什么不是这样？

Looking at memory before the program starts and before the scanf I would expect the memory to increase of 1 GB. Why isn't this happening?

编辑：我添加

byte[999999999] = 'a';
printf("%c",byte[999999999]);

程序输出 A 。

推荐答案

默认情况下，Linux的懒洋洋地分配物理内存，它访问的第一次。您对的malloc 通话将分配虚拟内存块大，但也有尚未映射到它的物理内存的任意页面。到未映射页第一访问将导致故障，这内核将通过分配和映射的物理存储器中的一个或多个页面处理

By default, Linux allocates physical memory lazily, the first time it's accessed. Your call to malloc will allocate a large block of virtual memory, but there are not yet any pages of physical memory mapped to it. The first access to an unmapped page will cause a fault, which the kernel will handle by allocating and mapping one or more pages of physical memory.

要分配所有的物理内存，你必须在每一页上访问至少一个字节;或者你可以绕过的malloc 键，直接进入操作系统的东西像

To allocate all the physical memory, you'll have to access at least one byte on every page; or you could bypass malloc and go straight to the operating system with something like

mmap(0, 1000000000, PROT_NONE, MAP_PRIVATE | MAP_ANONYMOUS | MAP_POPULATE, -1, 0);

请注意使用 MAP_POPULATE 的填充页表，即立即分配物理内存。根据该手册页，这仅适用于相当新版本的Linux。

Note the use of MAP_POPULATE to populate the page tables, i.e. to allocate physical memory, immediately. According to the manpage, this only works on fairly recent versions of Linux.

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