> From: Derek Price
> Sent: Tuesday, September 20, 2005 12:44
>
> Then you should be okay, though a few additional tests may be in order
> just to be sure (does the pointer look like it is from the same general
> data space as a pointer returned from malloc(1) in the same program?
> can you write a single byte to it without memory protection catching a
> fault? Can you free it without a fault?)
Yes.
Yes but the "free" function detects the write past the end and displays
a dialog box with "Abort", "Retry", "Ignore" options. Nothing odd occurs
when "free" isn't called. Looks clean to me.
Yes.
This code:
#include <stdio.h>
#include <malloc.h>
int main(int argc, char* argv[])
{
char *pm, *pr;
printf("Hello World!\n");
printf("argv = %p\n", argv);
printf("\n");
pm = malloc( 0 );
pr = realloc( 0, 0 );
printf("malloc( 0 ) = %p\n", pm);
printf("realloc( 0, 0 ) = %p\n", pr);
printf("*realloc( 0, 0 ) = 0x%02X\n", *pr);
printf("*realloc( 0, 0 ) = 2\n");
*pr = 2;
printf("*realloc( 0, 0 ) = 0x%02X\n", *pr);
printf("\n");
free(pm); pm = NULL;
free(pr); pr = NULL;
pm = malloc( 1 );
pr = realloc( 0, 0 );
printf("malloc( 1 ) = %p\n", pm);
printf("realloc( 0, 0 ) = %p\n", pr);
printf("*realloc( 0, 0 ) = 0x%02X\n", *pr);
printf("*realloc( 0, 0 ) = 3\n");
*pr = 3;
printf("*realloc( 0, 0 ) = 0x%02X\n", *pr);
printf("\n");
free(pm); pm = NULL;
free(pr); pr = NULL;
return 0;
}
Does this:
Hello World!
argv = 002F0EE0
malloc( 0 ) = 002F07A8
realloc( 0, 0 ) = 002F07E0
*realloc( 0, 0 ) = 0xFFFFFFFD
*realloc( 0, 0 ) = 2
*realloc( 0, 0 ) = 0x02
malloc( 1 ) = 002F07A8
realloc( 0, 0 ) = 002F07E0
*realloc( 0, 0 ) = 0xFFFFFFFD
*realloc( 0, 0 ) = 3
*realloc( 0, 0 ) = 0x03
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