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Each of the following two methods transforms an int to a 2's complement binary string. Is there any reason to prefer one to the other? George W. Cherry http://sdml.com    public static String intToBinary1( int value ) {       StringBuffer result = new StringBuffer(32);       for( int i = 0; i < 32; i++ ) {          if(value < 0)             result.append('1');          else {             result.append('0');          }          value = value << 1;       }//end for       return result.toString();    }//end intToBinary1()    public static String intToBinary2( int value ) {       StringBuffer result = new StringBuffer(                                                 );       for( int i = 0; i < 32; i++ ) {          if( (value & 1) == 1 )             result.setCharAt(31 - i, '1');          else {             result.setCharAt(31 - i, '0');          }          value = value 1;       }//end for       return result.toString();    }//end intToBinary2()

string. Is there any reason to prefer one to the other? George W. Cherry http://sdml.com    public static String intToBinary1( int value ) {       StringBuffer result = new StringBuffer(32);       for( int i = 0; i < 32; i++ ) {          if(value < 0)             result.append('1');          else {             result.append('0');          }          value = value << 1;       }//end for       return result.toString();    }//end intToBinary1()    public static String intToBinary2( int value ) {       StringBuffer result = new StringBuffer(                                                 );       for( int i = 0; i < 32; i++ ) {          if( (value & 1) == 1 )             result.setCharAt(31 - i, '1');          else {             result.setCharAt(31 - i, '0');          }          value = value 1;       }//end for       return result.toString();    }//end intToBinary2() Here's an idea. You tell us which you think is better, and why, then we will discuss your reply. We don't do homework here. You could learn a whole lot by compiling and testing the two methods. Have you considered this ingenious approach? BTW there is a one-line solution in Java for this conversion.

string. Is there any reason to prefer one to the other? George W. Cherry Paul Lutus <<Here's an idea. You tell us which you think is better, and why, then we will discuss your reply. We don't do homework here. Good idea, Paul. I prefer intToBinary1() to intToBinary2() because it is shorter, simpler, and runs a little faster. Alas, my question is not homework: I graduated from MIT many decades ago and retired from MIT awhile ago also. (I was Associate Director of the Draper laboratory. One of my jobs there was project manager of the Apollo Primary Navigation, Guidance, and Control System: MIT was the prime contractor to NASA for the PNGCS.) Now I'm just noodling around here and working on my fifth book. (Oh yes, I also push over dead trees in the Rachel Carson Wildlife Refuge in Maine, near where I mostly live.) Paul Lutus <<You could learn a whole lot by compiling and testing the two methods. Have you considered this ingenious approach? Another good idea, Paul. I had already done that, but I just added a benchmark of Mr. Babu Kalakrishnan's code, which ran faster than my similar algorithm, probably because he used a char[] and I used a StringBuffer. BTW, I like his code (intToBinary0) most of the three: it's shortest and fastest. intToBinary0() took 2690 milliseconds. Babu Kalakrishnan's code intToBinary1() took 4180 milliseconds. intToBinary2() took 4890 milliseconds. Paul Lutus <<BTW there is a one-line solution in Java for this conversion. If you don't mean Integer.toBinaryString(), I hope you'll post it. The three methods are below. George W. Cherry http://sdml.com  (book manu_script_ waiting for reviewers)    public static String intToBinary0( int value ) {       char[] result = new char[32];       for (int i = 0; i < 32; i = i + 1) {          result[i] = (value < 0) ? '1' : '0' ;          value = value << 1;       }       return new String(result);    }    public static String intToBinary1( int value ) {       StringBuffer result = new StringBuffer(32);       for( int i = 0; i < 32; i++ ) {          if(value < 0)             result.append('1');          else {             result.append('0');          }          value = value << 1;       }//end for       return result.toString();    }//end intToBinary1()    public static String intToBinary2( int value ) {       StringBuffer result = new StringBuffer(                                                 );       for( int i = 0; i < 32; i++ ) {          if( (value & 1) == 1 )             result.setCharAt(31 - i, '1');          else {             result.setCharAt(31 - i, '0');          }          value = value 1;       }//end for       return result.toString();    }//end intToBinary2()

Paul Lutus <<BTW there is a one-line solution in Java for this conversion. If you don't mean Integer.toBinaryString(), I hope you'll post it. How about with Long.toBinaryString()? ... Long.toBinaryString(((long) value & 0xffffffffl) | 0x100000000l).substring(1)

intToBinary0() took 2690 milliseconds. Babu Kalakrishnan's code intToBinary1() took 4180 milliseconds. intToBinary2() took 4890 milliseconds.   public static String intToBinary0( int value ) {      char[] result = new char[32];      for (int i = 0; i < 32; i = i + 1) {         result[i] = (value < 0) ? '1' : '0' ;         value = value << 1;      }      return new String(result);   } If you're looking at optimization, you might just be able to squeeze out a little bit more on with the following - at the cost of an extra couple of lines of code. public static String intToBinary3 (int value) {     char[] result = new char[32];     int i=0;     while (value != 0)     {         result[i++] = (value < 0) ? '1' : '0';         value = value << 1;     }     while (i < 32) result[i++]='0'; } I haven't benchmarked it - but this should save at least 1 shift on all even numbers, 2 shifts and 1 comparison for all multiples of 4, and so on (till for the single number 0 where it saves 32 shifts and 31 comparisons) at the cost of one extra integer comparison for all odd numbers. So on the average I guess it should be slightly faster over a uniformly distributed integer sample. BK