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Re: Memory leak in for loops


From: Chet Ramey
Subject: Re: Memory leak in for loops
Date: Mon, 04 May 2009 21:12:15 -0400
User-agent: Thunderbird 2.0.0.21 (Macintosh/20090302)

Mike Frysinger wrote:
> On Thursday 30 April 2009 05:56:02 Jan Schampera wrote:
>> Sandino Araico Sánchez wrote:
>>>    1.
>>>       #!/bin/bash
>>>    2.
>>>
>>>    3.
>>>       for i in {0..150000000} ; do
>>>    4.
>>>               echo $i > /dev/null
>>>    5.
>>>       done
>>>
>>>
>>>
>>> Repeat-By:
>>>         Run the script above and the process starts leaking memory very
>>> fast.
>> You know what a memory *leak* is, yes? mallocs() without proper free()s.
>>
>> What you mean is that your memory is used. Feel free to calculate the
>> memory that is needed to store the string representation of
>> {0..150000000}, I think you will get a number near your RAM size.
> 
> granted his example is wrong and his reasoning incorrect, there may actually 
> be a leak here ...
> 
> bash-4.0$ ps -p $$ -F
> UID        PID  PPID  C    SZ   RSS PSR STIME TTY          TIME CMD
> vapier   27999 16493  0  5675  3052   1 18:51 pts/1    00:00:00 bash --norc
> bash-4.0$ for n in {0..150000000} ; do echo $n; done
> <wait a few seconds>^C
> bash-4.0$ ps -p $$ -F
> UID        PID  PPID  C    SZ   RSS PSR STIME TTY          TIME CMD
> vapier   27999 16493 17 392694 473236 1 18:51 pts/1    00:00:01 bash --norc
> 
> that memory never goes away and bash will chew significant CPU (~10%) while 
> it 
> should be idle ...

That's not a memory leak.  Malloc implementations need not release
memory back to the kernel; the bash malloc (and others) do so only
under limited circumstances.  Memory obtained from the kernel using
mmap or sbrk and kept in a cache by a malloc implementation doesn't
constitute a leak.  A leak is memory for which there is no longer a
handle, by the application or by malloc itself.

Chet
-- 
``The lyf so short, the craft so long to lerne.'' - Chaucer

Chet Ramey, ITS, CWRU    address@hidden    http://cnswww.cns.cwru.edu/~chet/




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