From:	"Lars Hansen" <lars DOT o DOT hansen AT gmx DOT de>
Newsgroups:	comp.os.msdos.djgpp
Subject:	each address in memory "holds" 32 bit ?
Date:	Sat, 18 Jan 2003 22:29:29 +0100
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#include "stdio.h"
#include "string.h"

int main()
 {
  int i=10;
  int x[10]={0,1,2,3,4,5,6,7,8,9};
  memmove(x+2,x+3,sizeof(int)*7);
  while(i--) printf("%d,",x[i]);
 }

outputs  9,9,8,7,6,5,4,3,1,0

the same version with  memmove(x+sizeof(int)*2,x+sizeof(int)*3,sizeof(int)*7);

outputs  0,0,7,6,5,4,3,2,1,0


So is each address a 32 bit "place" or does the compiler "add" sizeof(int)*
automatically?
I had expect x to be an address (so having a range of an (long) int) and
adding an int increases the address by the value of the int so if the next
memory object I want is 2 bytes further away I have to add 2 ?

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