So malloc doesn't invoke any syscall?

问题

Related code:

  write(-1, "test", sizeof("test"));
  void * p = malloc(1024);
  void * p2 = malloc(510);
  write(-1, "hi", sizeof("hi"));

Related strace output:

write(4294967295, "test\0", 5)          = -1 EBADF (Bad file descriptor)
brk(0)                                  = 0x601000
brk(0x622000)                           = 0x622000
write(4294967295, "hi\0", 3)            = -1 EBADF (Bad file descriptor)

I'm surprised such low level operation doesn't involve syscall?

回答1:

Not every call to malloc invokes a syscall. On my linux desktop malloc allocates a space in 128KB blocks and then distributes the space. So I will see a syscall every 100-200 malloc calls. On freebsd malloc allocates by 2MB blocks. On your machine numbers will likely differ.

If you want to see syscall on every malloc allocate large amounts of memory (malloc(10*1024*1024*1024))

回答2:

What do you think brk is? malloc absolutely is invoking a syscall in this example, the syscall just isn't "malloc".

回答3:

malloc() calls the system brk() function (in Linux/Unix) - but it only calls it if the local heap is exhausted. I.e. most malloc implementations manage a memory heap obtained via brk(), and if it's too small or too fragmented they ask for more via brk().

来源：https://stackoverflow.com/questions/6326474/so-malloc-doesnt-invoke-any-syscall