If Windows lets you get away with this, then Windows is broken.
memset(ch,'\0',strlen(ch) );
'ch' is uninitialized local data. Nobody knows what evil lurks...
Thay said, the kernel will make sure that any data that gets
put into your address-space doesn't contain anybody else's
information --that's all. The junk on your stack was created
by your task.
On Thu, 27 Sep 2007, mahamuni ashish wrote:
> I have small code....
>
> #include <stdio.h>
> #include <string.h>
>
> int main()
> {
> float f= 1256.35;
> char ch[4];
>
> printf("\n1. f : %f",f);
> memset(ch,'\0',strlen(ch) );
> printf("\n2. f : %f",f);
> return 0;
> }
>
> Expected output is
> 1. f : 1256.35
> 2. f : 1256.35
>
> But I am getting the output
> (on windows)
> 1. f : 1256.35
> 2. f : 0.000000
>
> (on Linux)
> 1. f : 1256.35
> segmentation fault
>
> why?
>
>
> Forgot the famous last words? Access your message archive online at
> http://in.messenger.yahoo.com/webmessengerpromo.php
>
> -
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>
Cheers,
Dick Johnson
Penguin : Linux version 2.6.22.1 on an i686 machine (5588.29 BogoMips).
My book : http://www.AbominableFirebug.com/
_
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