Byte swap question (Bitwise operation) [on hold] The Next CEO of Stack Overflowsystem call to peek next byte from serial portTest question regarding grep

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Byte swap question (Bitwise operation) [on hold]



The Next CEO of Stack Overflowsystem call to peek next byte from serial portTest question regarding grep










-1















I have seen some bitwise operation practice questions (in C language) and there was one I didn't understand: you had to swap the nth and the mth byte in a specifc integer "x" using only bitwise operations, and that n and m were both <=3. The solution involved masking the last 8 bits of the number resulting from xor based on x >> (n<<3) to x >> (m<<3), but the part I don't understand is why they shifted m/n right 3.






c







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put on hold as off-topic by ilkkachu, Rui F Ribeiro, muru, jimmij, maulinglawns yesterday



  • This question does not appear to be about Unix or Linux within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.




















    -1















    I have seen some bitwise operation practice questions (in C language) and there was one I didn't understand: you had to swap the nth and the mth byte in a specifc integer "x" using only bitwise operations, and that n and m were both <=3. The solution involved masking the last 8 bits of the number resulting from xor based on x >> (n<<3) to x >> (m<<3), but the part I don't understand is why they shifted m/n right 3.






    c







    share|improve this question















    put on hold as off-topic by ilkkachu, Rui F Ribeiro, muru, jimmij, maulinglawns yesterday



    • This question does not appear to be about Unix or Linux within the scope defined in the help center.
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      -1












      -1








      -1








      I have seen some bitwise operation practice questions (in C language) and there was one I didn't understand: you had to swap the nth and the mth byte in a specifc integer "x" using only bitwise operations, and that n and m were both <=3. The solution involved masking the last 8 bits of the number resulting from xor based on x >> (n<<3) to x >> (m<<3), but the part I don't understand is why they shifted m/n right 3.






      c







      share|improve this question
















      I have seen some bitwise operation practice questions (in C language) and there was one I didn't understand: you had to swap the nth and the mth byte in a specifc integer "x" using only bitwise operations, and that n and m were both <=3. The solution involved masking the last 8 bits of the number resulting from xor based on x >> (n<<3) to x >> (m<<3), but the part I don't understand is why they shifted m/n right 3.






      c




      c






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 days ago









      Rui F Ribeiro

      41.8k1483142




      41.8k1483142










      asked 2 days ago









      Goodwin LuGoodwin Lu

      435




      435




      put on hold as off-topic by ilkkachu, Rui F Ribeiro, muru, jimmij, maulinglawns yesterday



      • This question does not appear to be about Unix or Linux within the scope defined in the help center.
      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by ilkkachu, Rui F Ribeiro, muru, jimmij, maulinglawns yesterday



      • This question does not appear to be about Unix or Linux within the scope defined in the help center.
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
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          Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (23 = 8).



          With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)



          (I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)






          share|improve this answer





























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (23 = 8).



            With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)



            (I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)






            share|improve this answer



























              1














              Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (23 = 8).



              With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)



              (I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)






              share|improve this answer

























                1












                1








                1







                Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (23 = 8).



                With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)



                (I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)






                share|improve this answer













                Because if n is a position within the word, in bytes, then 8*n is the same position in bits. And n<<3 is the same as 8*n (23 = 8).



                With n=1, n<<3 is 8, and (x >> (n << 3)) & 0xff would move the second lowest byte to the bottom position, and mask the rest of the word out. (The bytes would numbered starting from zero, of course.)



                (I'm assuming bytes of 8 bits, since I think that's by far the most common case at least nowadays.)







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 days ago









                ilkkachuilkkachu

                62.8k10103180




                62.8k10103180













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