Re: Re:expected calloc out put

[Sharyathi Nagesh <sharyath at in dot ibm dot com>]

I am using RHEL 5 Beta 1
kernel version 2.6.17-1.2519.4.21.el5

On Wed, 2006-10-25 at 09:59 -0500, Steven Munroe wrote:
> Sharyathi Nagesh <sharyath@in.ibm.com> writes:
> 
> >	ltp test suite tests for 2 successive calloc()s like this 
> >=====================
> >#include<stdio.h>
> >#include<stdlib.h>
> >#include<errno.h>
> >#define MAX 8092*8092
> >int main()
> >{
> >        char *ptr1,*ptr2;
> >        ptr1=calloc(MAX,1);
> >        perror("calloc 1:");
> >        ptr2=ptr1;
> >        free(ptr1);
> >        perror("free ");
> >        ptr1=calloc(MAX,1);
> >        perror(" calloc 2");
> >        if(ptr1 == ptr2)
> >                printf("\n Success");
> >        else
> >                printf("\n Failure");
> >}
> >=====================
> >fails in ppc 64 specifically. The test suite fails as ptr1 is not equal
> >to ptr2. Is calloc() expected to return same values on successive
> >allocation? as man page doesn't suggest any thing on this.
> 
> Your allocation is large enought to exceed the MMAP_THRESHOLD and use an
> annonymous mmap for the storage. In this case the free is just an unmap
> and the 2nd calloc is a new mmap. So it is the kernel choosing the address.
> 
> So what kernel and which distro are you running? RedHat? Fedora? 
> This might be execshield randomizing mmap addresses.
> 
>