What is a^b and (a&b)<<1?












8















I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is answer's JS:




var getSum=function(a,b){

         const Sum=a^b;//I can't understand it.Please give me an example to understand it

         const carry=(a&b)<<1;//I can't understand it too

               if(!carry){

                    return Sum

               }

        return getSum(Sum,carry);

};

console.log(getSum(5,1));








javascript







share|improve this question



























    8















    I was doing this question in leetcode.



    Request:




    Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




    I can't understand the solution it gave



    Could someone explain how this getSum function works?



    Here is answer's JS:




    var getSum=function(a,b){
    
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
    
         const carry=(a&b)<<1;//I can't understand it too
    
               if(!carry){
    
                    return Sum
    
               }
    
        return getSum(Sum,carry);
    
};
    
console.log(getSum(5,1));








    javascript







    share|improve this question

























      8












      8








      8


      1






      I was doing this question in leetcode.



      Request:




      Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




      I can't understand the solution it gave



      Could someone explain how this getSum function works?



      Here is answer's JS:




      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));








      javascript







      share|improve this question














      I was doing this question in leetcode.



      Request:




      Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




      I can't understand the solution it gave



      Could someone explain how this getSum function works?



      Here is answer's JS:




      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));








      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));





      var getSum=function(a,b){
      
         const Sum=a^b;//I can't understand it.Please give me an example to understand it
      
         const carry=(a&b)<<1;//I can't understand it too
      
               if(!carry){
      
                    return Sum
      
               }
      
        return getSum(Sum,carry);
      
};
      
console.log(getSum(5,1));





      javascript




      javascript






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      JackyJacky

      1758




      1758
























          3 Answers
          3






          active

          oldest

          votes


















          9














          Let's imagine that a = 3 and b = 5



          In binary notation they are a = 0011 and b = 0101



          XOR:
          a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



          So when we're doing a^b result will be 0110 (6 in decimal)



          AND + SHIFT



          a&b performs logical AND operation. It returns 1 only when a = b = 1.



          In our case the result is 0001



          << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



          Next iterations:



          Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



          Now a^b = 0100 and (a&b)<<1 = 0100



          Repeating again.



          Now a^b = 0000 and (a&b)<<1 = 1000



          And again.



          Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



          Everything worked fine since 3+5=8






          share|improve this answer





















          • 1





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            1 hour ago



















          1














           int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
          
int carry = (p & q) << 1; // Left Shift, 1+1=2
          
if (carry != 0) {
          
    return getSum(result, carry);
          
}
          
return result;


          Start By p=5,q=6. Then the XOR would be,



          0101
          
0110
          
------
          
0011


          So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



          0101
          
0110
          
-------
          
0100


          We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



           0100<<1=1000


          So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



          getSum(3, 8);


          So, doing the first XORing we get,



          0011
          
1000
          
-------
          
1011


          The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



          0011
          
1000
          
-------
          
0000


          We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
          As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



          Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






          share|improve this answer

































            -1














            These are bitwise operations. They're close to hardware language.






            share|improve this answer



















            • 7





              Too short.You didn't explain why this works and how

              – Jacky
              1 hour ago






            • 1





              Link-only answers are discouraged here. Please add the relevant content in the answer itself.

              – Ian McLaird
              1 hour ago






            • 3





              It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

              – Nate
              1 hour ago








            • 1





              @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

              – user202729
              34 mins ago






            • 2





              @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

              – user202729
              32 mins ago











            Your Answer






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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9














            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer





















            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              1 hour ago
















            9














            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer





















            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              1 hour ago














            9












            9








            9







            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8






            share|improve this answer















            Let's imagine that a = 3 and b = 5



            In binary notation they are a = 0011 and b = 0101



            XOR:
            a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



            So when we're doing a^b result will be 0110 (6 in decimal)



            AND + SHIFT



            a&b performs logical AND operation. It returns 1 only when a = b = 1.



            In our case the result is 0001



            << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



            Next iterations:



            Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



            Now a^b = 0100 and (a&b)<<1 = 0100



            Repeating again.



            Now a^b = 0000 and (a&b)<<1 = 1000



            And again.



            Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



            Everything worked fine since 3+5=8







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            vicodinvicodin

            1,097624




            1,097624








            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              1 hour ago














            • 1





              Great explanation! I always find the bitwise operations hard to understand

              – Francisco Hanna
              1 hour ago








            1




            1





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            1 hour ago





            Great explanation! I always find the bitwise operations hard to understand

            – Francisco Hanna
            1 hour ago













            1














             int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
            
int carry = (p & q) << 1; // Left Shift, 1+1=2
            
if (carry != 0) {
            
    return getSum(result, carry);
            
}
            
return result;


            Start By p=5,q=6. Then the XOR would be,



            0101
            
0110
            
------
            
0011


            So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



            0101
            
0110
            
-------
            
0100


            We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



             0100<<1=1000


            So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



            getSum(3, 8);


            So, doing the first XORing we get,



            0011
            
1000
            
-------
            
1011


            The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



            0011
            
1000
            
-------
            
0000


            We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
            As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



            Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






            share|improve this answer






























              1














               int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
              
int carry = (p & q) << 1; // Left Shift, 1+1=2
              
if (carry != 0) {
              
    return getSum(result, carry);
              
}
              
return result;


              Start By p=5,q=6. Then the XOR would be,



              0101
              
0110
              
------
              
0011


              So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



              0101
              
0110
              
-------
              
0100


              We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



               0100<<1=1000


              So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



              getSum(3, 8);


              So, doing the first XORing we get,



              0011
              
1000
              
-------
              
1011


              The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



              0011
              
1000
              
-------
              
0000


              We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
              As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



              Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






              share|improve this answer




























                1












                1








                1







                 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
                
int carry = (p & q) << 1; // Left Shift, 1+1=2
                
if (carry != 0) {
                
    return getSum(result, carry);
                
}
                
return result;


                Start By p=5,q=6. Then the XOR would be,



                0101
                
0110
                
------
                
0011


                So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



                0101
                
0110
                
-------
                
0100


                We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



                 0100<<1=1000


                So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



                getSum(3, 8);


                So, doing the first XORing we get,



                0011
                
1000
                
-------
                
1011


                The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



                0011
                
1000
                
-------
                
0000


                We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
                As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



                Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






                share|improve this answer















                 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
                
int carry = (p & q) << 1; // Left Shift, 1+1=2
                
if (carry != 0) {
                
    return getSum(result, carry);
                
}
                
return result;


                Start By p=5,q=6. Then the XOR would be,



                0101
                
0110
                
------
                
0011


                So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



                0101
                
0110
                
-------
                
0100


                We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



                 0100<<1=1000


                So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



                getSum(3, 8);


                So, doing the first XORing we get,



                0011
                
1000
                
-------
                
1011


                The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



                0011
                
1000
                
-------
                
0000


                We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
                As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



                Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                Ayan_84Ayan_84

                520513




                520513























                    -1














                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer



















                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      1 hour ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      1 hour ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      1 hour ago








                    • 1





                      @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                      – user202729
                      34 mins ago






                    • 2





                      @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                      – user202729
                      32 mins ago
















                    -1














                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer



















                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      1 hour ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      1 hour ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      1 hour ago








                    • 1





                      @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                      – user202729
                      34 mins ago






                    • 2





                      @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                      – user202729
                      32 mins ago














                    -1












                    -1








                    -1







                    These are bitwise operations. They're close to hardware language.






                    share|improve this answer













                    These are bitwise operations. They're close to hardware language.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 1 hour ago









                    WofWcaWofWca

                    40819




                    40819








                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      1 hour ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      1 hour ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      1 hour ago








                    • 1





                      @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                      – user202729
                      34 mins ago






                    • 2





                      @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                      – user202729
                      32 mins ago














                    • 7





                      Too short.You didn't explain why this works and how

                      – Jacky
                      1 hour ago






                    • 1





                      Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                      – Ian McLaird
                      1 hour ago






                    • 3





                      It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                      – Nate
                      1 hour ago








                    • 1





                      @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                      – user202729
                      34 mins ago






                    • 2





                      @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                      – user202729
                      32 mins ago








                    7




                    7





                    Too short.You didn't explain why this works and how

                    – Jacky
                    1 hour ago





                    Too short.You didn't explain why this works and how

                    – Jacky
                    1 hour ago




                    1




                    1





                    Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                    – Ian McLaird
                    1 hour ago





                    Link-only answers are discouraged here. Please add the relevant content in the answer itself.

                    – Ian McLaird
                    1 hour ago




                    3




                    3





                    It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                    – Nate
                    1 hour ago







                    It’s binary math. WofWca would have to give you a very very long explanation on binary to explain it. You should really read the W3 Schools article or watch some videos on Bitwise calculations. It’s the kind of thing that would be a few days of class work in college. The comment about link-only answers is fair, but I doubt you’ll grasp the concept from a StackOverflow post alone though.

                    – Nate
                    1 hour ago






                    1




                    1





                    @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                    – user202729
                    34 mins ago





                    @IanMcLaird It's not a link-only answer. The link content ("bitwise operations") is sufficient as an answer. Similar to an answer such as "Use the method abc.def.xyz".

                    – user202729
                    34 mins ago




                    2




                    2





                    @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                    – user202729
                    32 mins ago





                    @Jacky You didn't say that you don't know those, did you? If you had known what bitwise operations are, this answers would be sufficient (which may happen in case you are familiar with bitwise operations in another language, and don't know where to find the documentation and is completely unfamiliar with JS)

                    – user202729
                    32 mins ago


















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