Does malloc() use brk() or mmap()?

Question

c code:

// 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; } 

When compiling following code:

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

I get warning tip:

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

Question 1: Why is that?

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

Here is the output:

9b12000 9b12000 9b12000 

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?

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[update] Summary: brk() or mmap()

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

• 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;

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.

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

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

Answer 1

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; } 

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.

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.