how comes the two addresses equal while the digit are different

[khagesh gupta]

 
void main(){
 
int far *p=(int *)0X70230000;
 
int far *q=(int *)0XB0210000;
 
int near *x,near*y;
 
x=(int near *)p;
 
y=(int near *)q;
 
if(x==y)
 
printf("Both pointer are equal");
 
else
 
printf("Both pointer are not equal");
 
 
}
 

Output: Both pointers are equal

[weaknessforcats]

If I remove all that far/near stuff the pointers are not equal.

[donbock]

far and near are not part of Standard C, thus their operation is implementation-dependent.

In general, a near pointer value provides only the lower bits of the address; the upper bits are assumed. How many bits come from the near pointer value and how many are assumed could vary between implementations. In your case I suppose the break comes somewhere in the lowest 17 bits.

[Banfa]

As far (no pun intended) as I am aware near and far have only ever been used in 16 bit Windows.

In 32 bit windows I believe they are defined to nothing so on a 32 bit system that will produce "Both pointer are not equal" which is bourne out when I compile it with MinGW.

If you compiled it on a 16 bit windows platform then you must take account of the fact that near pointers were 16 bits and far pointers where 32 bits (to access memory in extended memory segments), consisting of a 16 bit segment and a 16 bit offset.

When you convert your far pointers to near pointers you drop the segment and are left with the offset (the lower 16 bits). Looking at your addresses the lower 16 bits of both of 0x0000 so under conversion to near pointers they are equal. Of course at least 1 if not both of them now no longer points to the right place either.