CS360 Lecture notes -- Malloc Lecture #1 -- note about different systems

  • Jim Plank
  • Directory: /blugreen/homes/plank/cs360/notes/Malloc1
  • Lecture notes: http://www.cs.utk.edu/~plank/plank/classes/cs360/360/notes/Malloc1/diff.html
    Each system implements malloc() in its own way. The description that I give in class is just one way of doing it, but the others are related. In the lecture notes, I give the following output for fb4: 
    UNIX> fb4
sbrk(0) = 0x70f8
Allocated 4 bytes.  buf = 0x61a8, buf[-1] = 0, buf[-2] = 16, buf[0] = 1000
Allocated 8 bytes.  buf = 0x61b8, buf[-1] = 0, buf[-2] = 16, buf[0] = 1001
Allocated 12 bytes.  buf = 0x61c8, buf[-1] = 0, buf[-2] = 24, buf[0] = 1002
Allocated 16 bytes.  buf = 0x61e0, buf[-1] = 0, buf[-2] = 24, buf[0] = 1003
Allocated 20 bytes.  buf = 0x61f8, buf[-1] = 0, buf[-2] = 32, buf[0] = 1004
Allocated 24 bytes.  buf = 0x6218, buf[-1] = 0, buf[-2] = 32, buf[0] = 1005
Allocated 28 bytes.  buf = 0x6238, buf[-1] = 0, buf[-2] = 40, buf[0] = 1006
Allocated 100 bytes.  buf = 0x6260, buf[-1] = 0, buf[-2] = 112, buf[0] = 1007
sbrk(0) = 0x70f8
UNIX>
    This was the output on one of the cetus machines in 1996. In 1999, I get the following output on cetus1a: 
    UNIX> fb4
sbrk(0) = 0x20b08
Allocated 4 bytes.  buf = 0x20b18, buf[-1] = 0, buf[-2] = 9, buf[0] = 1000
Allocated 8 bytes.  buf = 0x20b28, buf[-1] = 0, buf[-2] = 9, buf[0] = 1001
Allocated 12 bytes.  buf = 0x20f20, buf[-1] = 0, buf[-2] = 17, buf[0] = 1002
Allocated 16 bytes.  buf = 0x20f38, buf[-1] = 0, buf[-2] = 17, buf[0] = 1003
Allocated 20 bytes.  buf = 0x21528, buf[-1] = 0, buf[-2] = 25, buf[0] = 1004
Allocated 24 bytes.  buf = 0x21548, buf[-1] = 0, buf[-2] = 25, buf[0] = 1005
Allocated 28 bytes.  buf = 0x21d30, buf[-1] = 0, buf[-2] = 33, buf[0] = 1006
Allocated 100 bytes.  buf = 0x22730, buf[-1] = 0, buf[-2] = 105, buf[0] = 1007
sbrk(0) = 0x22b08
UNIX>
    This is different from the example output above, but as you can see, it is functionally equivalent -- the value in the word two words before the returned pointer is seven less than the number of bytes allocated. Why? I don't know -- but all you have to do is add seven to it to get.

    The Pentium box on my desk runs Linux 2.2.10, and it has the following output: 

    UNIX> fb4
sbrk(0) = 0x8049778
Allocated 4 bytes.  buf = 0x8049780, buf[-1] = 17, buf[-2] = 0, buf[0] = 1000
Allocated 8 bytes.  buf = 0x8049790, buf[-1] = 17, buf[-2] = 0, buf[0] = 1001
Allocated 12 bytes.  buf = 0x80497a0, buf[-1] = 17, buf[-2] = 0, buf[0] = 1002
Allocated 16 bytes.  buf = 0x80497b0, buf[-1] = 25, buf[-2] = 0, buf[0] = 1003
Allocated 20 bytes.  buf = 0x80497c8, buf[-1] = 25, buf[-2] = 0, buf[0] = 1004
Allocated 24 bytes.  buf = 0x80497e0, buf[-1] = 33, buf[-2] = 0, buf[0] = 1005
Allocated 28 bytes.  buf = 0x8049800, buf[-1] = 33, buf[-2] = 0, buf[0] = 1006
Allocated 100 bytes.  buf = 0x8049820, buf[-1] = 105, buf[-2] = 0, buf[0] = 1007
sbrk(0) = 0x804a000
    So now the size minus seven is in the first word before the pointer.

    Here's the output on a DECstation running Ultrix: 

    UNIX> fb4
sbrk(0) = 0x10001100
Allocated 4 bytes.  buf = 0x10005000, buf[-1] = 268435695, buf[-2] = 0, buf[0] = 1000
Allocated 8 bytes.  buf = 0x10006000, buf[-1] = 268435951, buf[-2] = 0, buf[0] = 1001
Allocated 12 bytes.  buf = 0x10006010, buf[-1] = 268435951, buf[-2] = 0, buf[0] = 1002
Allocated 16 bytes.  buf = 0x10007000, buf[-1] = 268436207, buf[-2] = 0, buf[0] = 1003
Allocated 20 bytes.  buf = 0x10007020, buf[-1] = 268436207, buf[-2] = 0, buf[0] = 1004
Allocated 24 bytes.  buf = 0x10007040, buf[-1] = 268436207, buf[-2] = 0, buf[0] = 1005
Allocated 28 bytes.  buf = 0x10007060, buf[-1] = 268436207, buf[-2] = 0, buf[0] = 1006
Allocated 100 bytes.  buf = 0x10008000, buf[-1] = 268436719, buf[-2] = 0, buf[0] = 1007
sbrk(0) = 0x10008ffc
UNIX>
    There's an odd one -- Any clues? When I turn the int's into shorts in fb5.c, I get: 
     
UNIX> fb5
sbrk(0) = 0x10001100
Allocated 4 bytes.  buf = 0x10005000, buf[-1] = 4096, buf[-2] = 239, buf[0] = 1000
Allocated 8 bytes.  buf = 0x10006000, buf[-1] = 4096, buf[-2] = 495, buf[0] = 1001
Allocated 12 bytes.  buf = 0x10006010, buf[-1] = 4096, buf[-2] = 495, buf[0] = 1002
Allocated 16 bytes.  buf = 0x10007000, buf[-1] = 4096, buf[-2] = 751, buf[0] = 1003
Allocated 20 bytes.  buf = 0x10007020, buf[-1] = 4096, buf[-2] = 751, buf[0] = 1004
Allocated 24 bytes.  buf = 0x10007040, buf[-1] = 4096, buf[-2] = 751, buf[0] = 1005
Allocated 28 bytes.  buf = 0x10007060, buf[-1] = 4096, buf[-2] = 751, buf[0] = 1006
Allocated 100 bytes.  buf = 0x10008000, buf[-1] = 4096, buf[-2] = 1263, buf[0] = 1007
sbrk(0) = 0x10008ffc
UNIX>
    Odd indeed. Can you guess at an explanation? I think I can, but it will take a little probing. Stay tuned....