malloc()使用brk()还是mmap()吗?

问题描述

c代码:

// program break mechanism
// TLPI exercise 7-1

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

void program_break_test() {
    printf("%10p\n", sbrk(0));

    char *bl = malloc(1024 * 1024);
    printf("%x\n", sbrk(0));

    free(bl);
    printf("%x\n", sbrk(0));

}

int main(int argc, char **argv) {
    program_break_test();
    return 0;
}

在编译以下代码时:

 printf("%10p\n", sbrk(0));

我得到警告提示:

format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int’

问题1:为什么?

在我malloc(1024 * 1024)之后，程序中断似乎没有改变.

And after I malloc(1024 * 1024), it seems the program break didn't change.

以下是输出:

9b12000
9b12000
9b12000

问题2::进程在启动时是否在堆上分配内存以备将来使用?还是编译器更改分配的时间点?否则，为什么?

Question 2: Does the process allocate memory on heap when start for future use? Or the compiler change the time point to allocate? Otherwise, why?

[更新]摘要:brk()或mmap()

在审查了TLPI并检查了手册页(在TLPI的作者的帮助下)之后，现在我了解了malloc()如何决定使用brk()或mmap()，如下所示:

After reviewing TLPI and check man page (with help from author of TLPI), now I understand how malloc() decide to use brk() or mmap(), as following:

mallopt()可以设置参数来控制malloc()的行为，并且通常有一个名为M_MMAP_THRESHOLD的参数:

mallopt() could set parameters to control behavior of malloc(), and there is a parameter named M_MMAP_THRESHOLD, in general:

• 如果请求的内存小于它，将使用brk()；
• 如果请求的内存大于或等于它，将使用mmap()；

• If requested memory is less than it, brk() will be used;
• If requested memory is larger than or equals to it, mmap() will be used;

参数的默认值是128kb(在我的系统上)，但是在我的测试程序中，我使用了1Mb，所以选择了mmap()，当我将请求的内存更改为32kb时，我看到brk()是使用.

The default value of the parameter is 128kb (on my system), but in my testing program I used 1Mb, so mmap() was chosen, when I changed requested memory to 32kb, I saw brk() would be used.

这本书在TLPI第147和1035页中提到过，但是我没有仔细阅读这一部分.

The book mentioned that in TLPI page 147 and 1035, but I didn't read carefully of that part.

该参数的详细信息可以在mallopt()的手册页中找到.

Detailed info of the parameter could be found in man page for mallopt().

推荐答案

如果我们更改程序以查看malloc的内存在哪里:

If we change the program to see where the malloc'd memory is:

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

void program_break_test() {
  printf("%10p\n", sbrk(0));

  char *bl = malloc(1024 * 1024);
  printf("%10p\n", sbrk(0));
  printf("malloc'd at: %10p\n", bl);

  free(bl);
  printf("%10p\n", sbrk(0));

}

int main(int argc, char **argv) {
  program_break_test();
  return 0;
}

sbrk不会改变可能更清楚一些. malloc给我们的记忆被映射到一个完全不同的位置.

It's perhaps a bit clearer that sbrk wouldn't change. The memory given to us by malloc is being mapped into a wildly different location.

您还可以在Linux上使用strace来查看进行了哪些系统调用，并发现malloc正在使用mmap进行分配.

You could also use strace on Linux to see what system calls are made, and find out that malloc is using mmap to perform the allocation.

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