Re: Over-riding GCC's builtin functions

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Hi Philip,

   Thanks. I will try it again tonight at my home. I don't have access
to a Linux / GCC right now in office.

-- Satish.BD



On Sun, Sep 6, 2009 at 10:51 PM, Philip
Herron<herron.philip@xxxxxxxxxxxxxx> wrote:
> Philip Herron wrote:
>>
>> bd satish wrote:
>>>
>>> Hello,
>>>
>>>    I wrote my own version of memchr( ) but looks like it is clashing
>>> with GCC's builtin function of same name. I'm not able to set any
>>> breakpoint.
>>> How should I over-ride this ?
>>>
>>> Here is the code I wrote:
>>> ----- memchr.cpp ----
>>> void* memchr(const void* str,  int ch, size_t count)
>>> {
>>>    unsigned char* buf = (unsigned char*)str;
>>>    unsigned char  c   = ((unsigned char)(ch));
>>>    register unsigned int i;
>>>
>>>    for(i=0; i<count; i++)
>>>    {
>>>        if(*buf == c)            return (void*)buf;
>>>        buf++;
>>>    }
>>>    return NULL;
>>> }
>>> I compiled the above like:
>>> g++ -c -g -Wall -ansi -nostdinc -nostdinc++ -fno-builtin
>>> fno-nonansi-builtins -Wshadow memchr.cpp
>>>
>>> And to test the above code, here is a simple main( )
>>> --- test_memchr.cpp ---
>>> int main( )
>>> {
>>>    const char msg[ ] = "Hello World";
>>>    char x; char *c=&x;
>>>
>>>    c = (char*)memchr(msg, 'X', 10);
>>>    return *c;
>>> }
>>>
>>> The above was compiled as
>>>
>>> g++  -g -Wall -nostdinc++ -nostdinc -fno-builtin test_memchr.cpp memchr.o
>>>
>>> The program gives a segmentation fault ! And I can't put a breakpoint
>>> even though "-g" is specified.
>>>
>>> My guess is that I'm somewhere calling GCC's in built memchr instead
>>> of mine. Can somebody help me where am I going wrong ?
>>>
>>> -- Satish.BD
>>>
>>
>> I just done this on my laptop seems to work fine:
>>
>> #include <stdio.h>
>> #include <stdlib.h>
>> #include <string.h>
>>
>> void* memchr( const void* str, int ch, size_t count )
>> {
>>  unsigned char  c= (unsigned char)ch;
>>  const char* dt= (const char*) str;
>>  unsigned int i= 0;
>>
>>  for( ; i< count; ++i )
>>   {
>>     if( dt[i] == c )              return (void*)dt;
>>   }
>>  return NULL;
>> }
>>
>> int main( )
>> {
>>   const char *msg= "Hello World";
>>   char *pch = (char*)memchr(msg, 'W', strlen(msg));
>>   if (pch!=NULL)
>>     printf ("'X' found at position %li.\n", pch-msg+1 );
>>   else
>>     printf ("'X' not found.\n");
>>
>>   return 0;
>> }
>>
>> just based it off your code, just compiled with gcc test.c, its not that
>> clean or safe in my opinion, too many casts i think.
>
> I think your problem was with your: "char x; char *c=&x; "
> You declared x without any initializer and assigned c the address of x that
> doesn't have anything. Which is a bit dangerous. Not sure i didn't bother
> debugging too much with your code i just got segfault's so i played with a
> 10 second hack :).
>
>


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