OS: Windows XP Home SP2
CPU/Ram: Mobile AMD Duron 4, 1GHz / 256MB
System Manufacturer: Compaq Presario 730us
Using windiag.exe to test my RAM I get a consistent error for the
following address: 0b5d9270. The system returns the wrong value, e.g.
expects value f7f7f7f7 and instead gets f7f7e7f7. This RAM is on the
motherboard and it's about as expensive to replace it than to buy a
new computer.
Is there any small program that could be written so as to
reserve/exclude this specific RAM address at boot so that it's not
used? Unfortunately I know little programming ...
Somebody suggested an approach via C that may have some of the
following lines (but didn't elaborate more). Does this approach have
any chance of success?
char far *pointer=(char far *) 0x0b5d9270; /* pointer to address*/
or one could point to the previous address 0x0b5d926f
and then try from there (not sure how because the following functions
transfer memory from pre-established places)
#include "malloc.h"
#include "stdlib.h"
#include "stdio.h
void main(void){
char far *pointer=(char far *) 0x0b5d9270; /* pointer to address*/
/* char huge *pointer; - original code that asks from a
pre-established place*/
if ((pointer=halloc(1,128000))==NULL {
printf("allocation error -aborting");
exit(1);
}
/* do not free memory*/
/* hfree(pointer); */
}
Program should be called and left resident as a service or driver
during boot.
THANK YOU very much for your response, I really appreciate it.
Juan
