Sponsored Content
Top Forums Programming c program to set the m-bit to n-bit Post 302601636 by acdc on Friday 24th of February 2012 05:56:03 AM
Old 02-24-2012
c program to set the m-bit to n-bit

I have a 32bit number and without using for loop,I want to set mbit to n bit.

Say m bit may be 2nd or 5th or 9th or 10th.n bit may be 22nd or 27or 11th bit.

I assume m<n.

Please help me.Thanks

acdc
 

5 More Discussions You Might Find Interesting

1. Programming

copying or concatinating string from 1st bit, leaving 0th bit

Hello, If i have 2 strings str1 and str2, i would like to copy/concatenate str2 to str1, from 1st bit leaving the 0th bit. How do i do it? (2 Replies)
Discussion started by: jazz
2 Replies

2. Red Hat

boot the 32 bit kernel on a 64 bit PPC Linux machine?

Hi all, I'm looking to cover a corner case for an upcoming test cycle. Is there a way to boot a RedHat Advanced Server 4 (update 3) installed on a Power PC machine to use a 32 bit kernel? This would be similar to what is done here -> https://www.unix.com/aix/26204-aix-platform.html I've done... (0 Replies)
Discussion started by: philrau
0 Replies

3. UNIX for Advanced & Expert Users

migrating unix mp-ras 32 bit to linux suse 64 bit

Hi. I need to migrate the whole unix environment from a Unix mp-ras 32 bit to a Linux Suse 64 bit. 1) can i use cpio to copy the data? 2) can i just copy the users from unix to linux or do i have to create them by hand 3) are there any other concerns i should worry about? thanx (1 Reply)
Discussion started by: mrodrig
1 Replies

4. Shell Programming and Scripting

How to handle 64 bit arithmetic operation at 32 bit compiled perl interpreter?H

Hi, Here is the issue. From the program snippet I have Base: 0x1800000000, Size: 0x3FFE7FFFFFFFF which are of 40 and 56 bits. SO I used use bignum to do the math but summing them up I always failed having correct result. perl interpreter info, perl, v5.8.8 built for... (0 Replies)
Discussion started by: rrd1986
0 Replies

5. Windows & DOS: Issues & Discussions

Which version of Windows Vista to install with a product key? 32-bit or 64-bit?

Hello everyone. I bought a dell laptop (XPS M1330) online which came without a hard drive. There is a Windows Vista Ultimate OEMAct sticker with product key at the bottom case. I checked dell website (here) for this model and it says this model supports both 32 and 64-bit version of Windows... (4 Replies)
Discussion started by: milhan
4 Replies
BITMAP_ONTO(9)						  Basic Kernel Library Functions					    BITMAP_ONTO(9)

NAME
bitmap_onto - translate one bitmap relative to another SYNOPSIS
void bitmap_onto(unsigned long * dst, const unsigned long * orig, const unsigned long * relmap, int bits); ARGUMENTS
dst resulting translated bitmap orig original untranslated bitmap relmap bitmap relative to which translated bits number of bits in each of these bitmaps DESCRIPTION
Set the n-th bit of dst iff there exists some m such that the n-th bit of relmap is set, the m-th bit of orig is set, and the n-th bit of relmap is also the m-th _set_ bit of relmap. (If you understood the previous sentence the first time your read it, you're overqualified for your current job.) In other words, orig is mapped onto (surjectively) dst, using the the map { <n, m> | the n-th bit of relmap is the m-th set bit of relmap }. Any set bits in orig above bit number W, where W is the weight of (number of set bits in) relmap are mapped nowhere. In particular, if for all bits m set in orig, m >= W, then dst will end up empty. In situations where the possibility of such an empty result is not desired, one way to avoid it is to use the bitmap_fold operator, below, to first fold the orig bitmap over itself so that all its set bits x are in the range 0 <= x < W. The bitmap_fold operator does this by setting the bit (m % W) in dst, for each bit (m) set in orig. Example [1] for bitmap_onto: Let's say relmap has bits 30-39 set, and orig has bits 1, 3, 5, 7, 9 and 11 set. Then on return from this routine, dst will have bits 31, 33, 35, 37 and 39 set. When bit 0 is set in orig, it means turn on the bit in dst corresponding to whatever is the first bit (if any) that is turned on in relmap. Since bit 0 was off in the above example, we leave off that bit (bit 30) in dst. When bit 1 is set in orig (as in the above example), it means turn on the bit in dst corresponding to whatever is the second bit that is turned on in relmap. The second bit in relmap that was turned on in the above example was bit 31, so we turned on bit 31 in dst. Similarly, we turned on bits 33, 35, 37 and 39 in dst, because they were the 4th, 6th, 8th and 10th set bits set in relmap, and the 4th, 6th, 8th and 10th bits of orig (i.e. bits 3, 5, 7 and 9) were also set. When bit 11 is set in orig, it means turn on the bit in dst corresponding to whatever is the twelth bit that is turned on in relmap. In the above example, there were only ten bits turned on in relmap (30..39), so that bit 11 was set in orig had no affect on dst. Example [2] for bitmap_fold + bitmap_onto: Let's say relmap has these ten bits set: 40 41 42 43 45 48 53 61 74 95 (for the curious, that's 40 plus the first ten terms of the Fibonacci sequence.) Further lets say we use the following code, invoking bitmap_fold then bitmap_onto, as suggested above to avoid the possitility of an empty dst result: unsigned long *tmp; // a temporary bitmap's bits bitmap_fold(tmp, orig, bitmap_weight(relmap, bits), bits); bitmap_onto(dst, tmp, relmap, bits); Then this table shows what various values of dst would be, for various orig's. I list the zero-based positions of each set bit. The tmp column shows the intermediate result, as computed by using bitmap_fold to fold the orig bitmap modulo ten (the weight of relmap). orig tmp dst 0 0 40 1 1 41 9 9 95 10 0 40 (*) 1 3 5 7 1 3 5 7 41 43 48 61 0 1 2 3 4 0 1 2 3 4 40 41 42 43 45 0 9 18 27 0 9 8 7 40 61 74 95 0 10 20 30 0 40 0 11 22 33 0 1 2 3 40 41 42 43 0 12 24 36 0 2 4 6 40 42 45 53 78 102 211 1 2 8 41 42 74 (*) (*) For these marked lines, if we hadn't first done bitmap_fold into tmp, then the dst result would have been empty. If either of orig or relmap is empty (no set bits), then dst will be returned empty. If (as explained above) the only set bits in orig are in positions m where m >= W, (where W is the weight of relmap) then dst will once again be returned empty. All bits in dst not set by the above rule are cleared. COPYRIGHT
Kernel Hackers Manual 2.6. July 2010 BITMAP_ONTO(9)
All times are GMT -4. The time now is 03:15 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy