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 49 mins ago









      JackyJacky

      1758




      1758
























          3 Answers
          3






          active

          oldest

          votes


















          8














          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





















          • 2





            Thanks for your explanation!

            – Jacky
            12 mins ago











          • You're welcome!

            – vicodin
            11 mins ago






          • 1





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

            – Francisco Hanna
            6 mins 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
              41 mins ago






            • 1





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

              – Ian McLaird
              34 mins 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
              34 mins ago













            • @Nate Sorry,But I am not the computer science's student

              – Jacky
              31 mins ago











            Your Answer






            StackExchange.ifUsing("editor", function () {
            StackExchange.using("externalEditor", function () {
            StackExchange.using("snippets", function () {
            StackExchange.snippets.init();
            });
            });
            }, "code-snippets");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "1"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55193135%2fwhat-is-ab-and-ab1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8














            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





















            • 2





              Thanks for your explanation!

              – Jacky
              12 mins ago











            • You're welcome!

              – vicodin
              11 mins ago






            • 1





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

              – Francisco Hanna
              6 mins ago
















            8














            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





















            • 2





              Thanks for your explanation!

              – Jacky
              12 mins ago











            • You're welcome!

              – vicodin
              11 mins ago






            • 1





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

              – Francisco Hanna
              6 mins ago














            8












            8








            8







            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 12 mins ago

























            answered 29 mins ago









            vicodinvicodin

            1,088624




            1,088624








            • 2





              Thanks for your explanation!

              – Jacky
              12 mins ago











            • You're welcome!

              – vicodin
              11 mins ago






            • 1





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

              – Francisco Hanna
              6 mins ago














            • 2





              Thanks for your explanation!

              – Jacky
              12 mins ago











            • You're welcome!

              – vicodin
              11 mins ago






            • 1





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

              – Francisco Hanna
              6 mins ago








            2




            2





            Thanks for your explanation!

            – Jacky
            12 mins ago





            Thanks for your explanation!

            – Jacky
            12 mins ago













            You're welcome!

            – vicodin
            11 mins ago





            You're welcome!

            – vicodin
            11 mins ago




            1




            1





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

            – Francisco Hanna
            6 mins ago





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

            – Francisco Hanna
            6 mins 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 14 mins ago

























                answered 19 mins 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
                      41 mins ago






                    • 1





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

                      – Ian McLaird
                      34 mins 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
                      34 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      31 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
                      41 mins ago






                    • 1





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

                      – Ian McLaird
                      34 mins 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
                      34 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      31 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 45 mins ago









                    WofWcaWofWca

                    40819




                    40819








                    • 7





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

                      – Jacky
                      41 mins ago






                    • 1





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

                      – Ian McLaird
                      34 mins 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
                      34 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      31 mins ago














                    • 7





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

                      – Jacky
                      41 mins ago






                    • 1





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

                      – Ian McLaird
                      34 mins 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
                      34 mins ago













                    • @Nate Sorry,But I am not the computer science's student

                      – Jacky
                      31 mins ago








                    7




                    7





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

                    – Jacky
                    41 mins ago





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

                    – Jacky
                    41 mins ago




                    1




                    1





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

                    – Ian McLaird
                    34 mins ago





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

                    – Ian McLaird
                    34 mins 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
                    34 mins 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
                    34 mins ago















                    @Nate Sorry,But I am not the computer science's student

                    – Jacky
                    31 mins ago





                    @Nate Sorry,But I am not the computer science's student

                    – Jacky
                    31 mins ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Stack Overflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55193135%2fwhat-is-ab-and-ab1%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Сан-Квентин

                    Image
                    Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
                    Read more

                    8-я гвардейская общевойсковая армия

                    Image
                    8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
                    Read more

                    Алькесар

                    Image
                    Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
                    Read more