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Thread: How to write....**
 08062008, 01:53 PM #1Member
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 08062008, 02:19 PM #2
Is this an assignment? Have you written some code you need help with?
 08062008, 02:38 PM #3Member
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 08062008, 05:28 PM #4Member
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I can see ur using recursing
where is the break condition?visit http://www.thejavacode.com
Muhammad Safwat
 08062008, 06:15 PM #5
Here's cod to do it bit by bit.
Java Code:// Add two ints using bitwise operators public class ArithWithBitOps { // // Second attempt  Go bit by bit right to left doing binary arith static int carry = 0; // c=1 for 1+1 static int add2(int a, int b) { int sum = 0; // get sum here int mask = 1; // shifting mask used to test each bit while(mask != 0) { int ta = a & mask; // get next bit to add int tb = b & mask; // get next bit to add int bsum = bitAdd2(ta, tb, mask); // add the bits & set carry if(Testing) System.out.print("ta=" + ta + ", tb=" + tb + ", bsum=" + bsum + ", carry=" + carry + ", b4sum=" + sum); sum = (sum  bsum); // OR in the results if(Testing) System.out.println(", aft sum=" + sum); mask = mask << 1; // move to next bit } // end while() thru bits if(carry > 0) System.err.println(">>>>>>>>>>>>losing carry"); return sum; } // end add2() // // Add two selected bits in x and y & set carry if carry static int bitAdd2(int x, int y, int bsel) { // System.out.println("bitAdd2 of " + x + " " + y); int tx = x & bsel; int ty = y & bsel; if(carry == 1) { // Have carry ? if((tx & ty) != 0) { carry = 1; return bsel; // 1 + 1 + c=1 > 1 + c=1 }else if((tx ^ ty) != 0) { carry = 1; return 0; // 1 + 0 + c=1 > 0 + c=1 }else { carry = 0; return bsel; // 0 + 0 + c=1 > 1 + c=0 } }else if (carry == 0) { // no carry if((tx & ty) > 0) { carry = 1; return 0; // 1 + 1 + c=0 > 0 + c=1 }else if((tx ^ ty) != 0) { carry = 0; return bsel; // 1 + 0 + c=0 > 1 + c=0 }else { System.out.println("(tx ^ ty)=" + (tx ^ ty)); carry = 0; return 0; // 0 + 0 + c=0 > 0 + c=0 } }else { System.err.println("Invalid carry= " + carry); return 0; } } // end bitAdd2() // // Test the above public static void main(String[] args) { // The numbers to add int x = 24; int y = 15; if(Testing) System.out.println("Max int=" + Integer.MAX_VALUE //Max int=2147483647 + " " + Integer.toHexString(2) //2=fffffffe + " " + 0x80000000); //2147483648 int sum = add2(x, y); System.out.println("Final sum= " + sum + " vs " + (x + y) + " " + Integer.toHexString(sum)); //Final sum= 9 vs 9 fffffff7 } // end main() static final boolean Testing = false; // global flag for debug output }
Last edited by Norm; 08062008 at 08:11 PM.
 08072008, 08:44 AM #6
Wow norm...
Good job...To finish sooner, take your own time....
Nivedithaaaa
 08072008, 10:00 AM #7Member
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Wonderful.... NORM....
Thanks vey much.............
 08072008, 01:56 PM #8
I don't see how you could answer this question in a interview. My version took a couple of hours.
 08072008, 02:06 PM #9Member
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 08072008, 02:09 PM #10Member
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Great Effort you have put in....
It was asked to my friend.. in Cognizant Technology ...Company... they just give him to slove this problem in 45 minutes... but he couldn't... if he does the problem he got he job....
Any Way.......... I got the code... but it is really unusefull to write such a code for simple addition....
Thanks a lot..........
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