Member Avatar for gerard4143

Is this the proper way to zero the first 12 bits of an address?

void *ans = (void*)((unsigned long)addr &  ((0UL - 1)  ^ 0xfff));

Where addr is any user process memory address..

Note this method works...I just want to know if there is a better way

Edited by gerard4143 because: n/a
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Member Avatar for kolosick.m188

I'd just do a bitwise shift left 12 then a bitwise shift right 12

Member Avatar for gerard4143

I'd just do a bitwise shift left 12 then a bitwise shift right 12

Don't you mean the other way around? Maybe I have a different meaning for first 12

Edited by gerard4143 because: n/a
Member Avatar for kolosick.m188

Here's what I mean
1010101010101111
First twelve is 101010101010
Shift left twelve gives
1111000000000000
Shift right twelve then gives
0000000000001111

Member Avatar for gerard4143

Here's what I mean
1010101010101111
First twelve is 101010101010
Shift left twelve gives
1111000000000000
Shift right twelve then gives
0000000000001111

O.K. what I meant was the opposite to what you have here...either way your method works..Thanks

Member Avatar for Dave Sinkula

Is this the proper way to zero the first 12 bits of an address?

void *ans = (void*)((unsigned long)addr &  ((0UL - 1)  ^ 0xfff));

Where addr is any user process memory address..

Note this method works...I just want to know if there is a better way

That kind of masking works just fine; it might be expressed more simply:

void *ans = (void*)((unsigned long)addr &  ~0xfffUL);

(At least as far as manipulating bits of an integer. I don't know what address manipulation will get ya.)

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