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C Language - add products’ digits (i.e., not the products themselves)



2019 Community Moderator ElectionGetting each individual digit from a whole integerImprove INSERT-per-second performance of SQLite?Is 'switch' faster than 'if'?How do I achieve the theoretical maximum of 4 FLOPs per cycle?execution time in C languageCan code that is valid in both C and C++ produce different behavior when compiled in each language?What is the fastest way to calculate e to 2 trillion digits?2 digit sums and all combinations for each sumFind the sum of digits of a number(in c)Limiting my float to only 3 DIGITSinteger overflow in expression, when multiplying 13 digits










0















I am looking to calculate the below sum but using the products digits, not the products themselves:



These are the initial values:



2 + 12 + 8 = 22



but what I want to achieve is the following, so that the digit 12 is actually seen as a 1 and a 2 separately



2 + 1 + 2 + 8 = 13



Using C language is there a formula in which I can use to perform this task?










share|improve this question

















  • 2





    No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

    – 4386427
    Mar 7 at 8:03






  • 1





    where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

    – bruno
    Mar 7 at 8:04












  • See stackoverflow.com/questions/3118490/…

    – 4386427
    Mar 7 at 8:06











  • Do you ever have any other operator between the numbers other than +?

    – Bathsheba
    Mar 7 at 8:10















0















I am looking to calculate the below sum but using the products digits, not the products themselves:



These are the initial values:



2 + 12 + 8 = 22



but what I want to achieve is the following, so that the digit 12 is actually seen as a 1 and a 2 separately



2 + 1 + 2 + 8 = 13



Using C language is there a formula in which I can use to perform this task?










share|improve this question

















  • 2





    No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

    – 4386427
    Mar 7 at 8:03






  • 1





    where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

    – bruno
    Mar 7 at 8:04












  • See stackoverflow.com/questions/3118490/…

    – 4386427
    Mar 7 at 8:06











  • Do you ever have any other operator between the numbers other than +?

    – Bathsheba
    Mar 7 at 8:10













0












0








0








I am looking to calculate the below sum but using the products digits, not the products themselves:



These are the initial values:



2 + 12 + 8 = 22



but what I want to achieve is the following, so that the digit 12 is actually seen as a 1 and a 2 separately



2 + 1 + 2 + 8 = 13



Using C language is there a formula in which I can use to perform this task?










share|improve this question














I am looking to calculate the below sum but using the products digits, not the products themselves:



These are the initial values:



2 + 12 + 8 = 22



but what I want to achieve is the following, so that the digit 12 is actually seen as a 1 and a 2 separately



2 + 1 + 2 + 8 = 13



Using C language is there a formula in which I can use to perform this task?







c






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 7 at 8:00









CoderCoder

9911




9911







  • 2





    No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

    – 4386427
    Mar 7 at 8:03






  • 1





    where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

    – bruno
    Mar 7 at 8:04












  • See stackoverflow.com/questions/3118490/…

    – 4386427
    Mar 7 at 8:06











  • Do you ever have any other operator between the numbers other than +?

    – Bathsheba
    Mar 7 at 8:10












  • 2





    No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

    – 4386427
    Mar 7 at 8:03






  • 1





    where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

    – bruno
    Mar 7 at 8:04












  • See stackoverflow.com/questions/3118490/…

    – 4386427
    Mar 7 at 8:06











  • Do you ever have any other operator between the numbers other than +?

    – Bathsheba
    Mar 7 at 8:10







2




2





No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

– 4386427
Mar 7 at 8:03





No, there is no standard C functions for doing that. However, it's easy to write some code doing that. In a loop you use digit = n % 10 to get a digit and n = n / 10 to prepare for getting the next digit.

– 4386427
Mar 7 at 8:03




1




1





where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

– bruno
Mar 7 at 8:04






where the values come from ? do you have them in an array for instance ? or do you have their external representation (string rather than int) ?

– bruno
Mar 7 at 8:04














See stackoverflow.com/questions/3118490/…

– 4386427
Mar 7 at 8:06





See stackoverflow.com/questions/3118490/…

– 4386427
Mar 7 at 8:06













Do you ever have any other operator between the numbers other than +?

– Bathsheba
Mar 7 at 8:10





Do you ever have any other operator between the numbers other than +?

– Bathsheba
Mar 7 at 8:10












3 Answers
3






active

oldest

votes


















1














Supposing you have an array of the values you can do :



#include <stdio.h>

unsigned sum(const unsigned * a, size_t sz)

unsigned sum = 0;

while (sz--)
unsigned v = *a++;

while (v)
sum += v%10;
v /= 10;



return sum;


int main()

const unsigned a[] = 2, 12, 8 ;

printf("sum = %un", sum(a, sizeof(a)/sizeof(*a)));



Compilation and execution :



/tmp % gcc -pedantic -Wextra s.c
/tmp % ./a.out
sum = 13





share|improve this answer






























    1














    If + is the only other token, then you can disregard it and merely sum the digits on the stream. So



    #include <stdio.h>
    #include <ctype.h> // for isdigit

    int main(void)

    char* s = "2 + 12 + 8";
    int total = 0;
    for (; *s; ++s)
    if (isdigit((unsigned char)*s))
    total += *s - '0';


    printf("%dn", total);



    is one way. *s - '0'; is the idiomatic way of transforming a char digit to its numerical value. The loop termination condition is the NUL terminator in the string s.



    This is most certainly not the best approach if you want other operators between terms. At that point you need to build a full expression parser (such as one based on the worked example in Kernighan & Ritchie).






    share|improve this answer




















    • 1





      To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

      – chqrlie
      Mar 7 at 8:22











    • @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

      – Bathsheba
      Mar 7 at 8:23


















    -1














    There is no direct formula for this task but you can perform this task in a separate function and use that function wherever it is needed.
    Here are two simple solutions which depend on the type of input.



    If Input is in the form of String (i.e. "2 + 12 + 8") then the code will be -



    #include <stdio.h>

    int main()

    char str[] = "2 + 12 + 8";
    int sum=0;

    for (int i = 0; i < strlen(str); i++)
    if (str[i] >= '0' && str[i] <= '9')
    sum += str[i] - '0';

    printf("%d",sum);
    return 0;



    If Input is in the form of Array (i.e. [2, 12, 8]) then the code will be -



    #include <stdio.h>

    int main()

    int num[] = 2, 12, 8;
    int sum=0;
    int length = sizeof(num) / sizeof(num[0]);

    for (int i = 0; i < length; i++)
    while (num[i])
    sum += num[i] % 10;
    num[i] /= 10;


    printf("%d",sum);
    return 0;






    share|improve this answer






















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      Supposing you have an array of the values you can do :



      #include <stdio.h>

      unsigned sum(const unsigned * a, size_t sz)

      unsigned sum = 0;

      while (sz--)
      unsigned v = *a++;

      while (v)
      sum += v%10;
      v /= 10;



      return sum;


      int main()

      const unsigned a[] = 2, 12, 8 ;

      printf("sum = %un", sum(a, sizeof(a)/sizeof(*a)));



      Compilation and execution :



      /tmp % gcc -pedantic -Wextra s.c
      /tmp % ./a.out
      sum = 13





      share|improve this answer



























        1














        Supposing you have an array of the values you can do :



        #include <stdio.h>

        unsigned sum(const unsigned * a, size_t sz)

        unsigned sum = 0;

        while (sz--)
        unsigned v = *a++;

        while (v)
        sum += v%10;
        v /= 10;



        return sum;


        int main()

        const unsigned a[] = 2, 12, 8 ;

        printf("sum = %un", sum(a, sizeof(a)/sizeof(*a)));



        Compilation and execution :



        /tmp % gcc -pedantic -Wextra s.c
        /tmp % ./a.out
        sum = 13





        share|improve this answer

























          1












          1








          1







          Supposing you have an array of the values you can do :



          #include <stdio.h>

          unsigned sum(const unsigned * a, size_t sz)

          unsigned sum = 0;

          while (sz--)
          unsigned v = *a++;

          while (v)
          sum += v%10;
          v /= 10;



          return sum;


          int main()

          const unsigned a[] = 2, 12, 8 ;

          printf("sum = %un", sum(a, sizeof(a)/sizeof(*a)));



          Compilation and execution :



          /tmp % gcc -pedantic -Wextra s.c
          /tmp % ./a.out
          sum = 13





          share|improve this answer













          Supposing you have an array of the values you can do :



          #include <stdio.h>

          unsigned sum(const unsigned * a, size_t sz)

          unsigned sum = 0;

          while (sz--)
          unsigned v = *a++;

          while (v)
          sum += v%10;
          v /= 10;



          return sum;


          int main()

          const unsigned a[] = 2, 12, 8 ;

          printf("sum = %un", sum(a, sizeof(a)/sizeof(*a)));



          Compilation and execution :



          /tmp % gcc -pedantic -Wextra s.c
          /tmp % ./a.out
          sum = 13






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Mar 7 at 8:11









          brunobruno

          10.2k21126




          10.2k21126























              1














              If + is the only other token, then you can disregard it and merely sum the digits on the stream. So



              #include <stdio.h>
              #include <ctype.h> // for isdigit

              int main(void)

              char* s = "2 + 12 + 8";
              int total = 0;
              for (; *s; ++s)
              if (isdigit((unsigned char)*s))
              total += *s - '0';


              printf("%dn", total);



              is one way. *s - '0'; is the idiomatic way of transforming a char digit to its numerical value. The loop termination condition is the NUL terminator in the string s.



              This is most certainly not the best approach if you want other operators between terms. At that point you need to build a full expression parser (such as one based on the worked example in Kernighan & Ritchie).






              share|improve this answer




















              • 1





                To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

                – chqrlie
                Mar 7 at 8:22











              • @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

                – Bathsheba
                Mar 7 at 8:23















              1














              If + is the only other token, then you can disregard it and merely sum the digits on the stream. So



              #include <stdio.h>
              #include <ctype.h> // for isdigit

              int main(void)

              char* s = "2 + 12 + 8";
              int total = 0;
              for (; *s; ++s)
              if (isdigit((unsigned char)*s))
              total += *s - '0';


              printf("%dn", total);



              is one way. *s - '0'; is the idiomatic way of transforming a char digit to its numerical value. The loop termination condition is the NUL terminator in the string s.



              This is most certainly not the best approach if you want other operators between terms. At that point you need to build a full expression parser (such as one based on the worked example in Kernighan & Ritchie).






              share|improve this answer




















              • 1





                To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

                – chqrlie
                Mar 7 at 8:22











              • @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

                – Bathsheba
                Mar 7 at 8:23













              1












              1








              1







              If + is the only other token, then you can disregard it and merely sum the digits on the stream. So



              #include <stdio.h>
              #include <ctype.h> // for isdigit

              int main(void)

              char* s = "2 + 12 + 8";
              int total = 0;
              for (; *s; ++s)
              if (isdigit((unsigned char)*s))
              total += *s - '0';


              printf("%dn", total);



              is one way. *s - '0'; is the idiomatic way of transforming a char digit to its numerical value. The loop termination condition is the NUL terminator in the string s.



              This is most certainly not the best approach if you want other operators between terms. At that point you need to build a full expression parser (such as one based on the worked example in Kernighan & Ritchie).






              share|improve this answer















              If + is the only other token, then you can disregard it and merely sum the digits on the stream. So



              #include <stdio.h>
              #include <ctype.h> // for isdigit

              int main(void)

              char* s = "2 + 12 + 8";
              int total = 0;
              for (; *s; ++s)
              if (isdigit((unsigned char)*s))
              total += *s - '0';


              printf("%dn", total);



              is one way. *s - '0'; is the idiomatic way of transforming a char digit to its numerical value. The loop termination condition is the NUL terminator in the string s.



              This is most certainly not the best approach if you want other operators between terms. At that point you need to build a full expression parser (such as one based on the worked example in Kernighan & Ritchie).







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Mar 7 at 8:30

























              answered Mar 7 at 8:13









              BathshebaBathsheba

              180k27254383




              180k27254383







              • 1





                To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

                – chqrlie
                Mar 7 at 8:22











              • @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

                – Bathsheba
                Mar 7 at 8:23












              • 1





                To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

                – chqrlie
                Mar 7 at 8:22











              • @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

                – Bathsheba
                Mar 7 at 8:23







              1




              1





              To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

              – chqrlie
              Mar 7 at 8:22





              To avoid undefined behavior on negative char values on systems that have signed char by default, you should write isdigit((unsigned char)*s). I might be more readable to write (*s >= '0' && *s <= '9')

              – chqrlie
              Mar 7 at 8:22













              @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

              – Bathsheba
              Mar 7 at 8:23





              @chqrlie: I can still here my professor shouting at me for this all those years ago. Fixed.

              – Bathsheba
              Mar 7 at 8:23











              -1














              There is no direct formula for this task but you can perform this task in a separate function and use that function wherever it is needed.
              Here are two simple solutions which depend on the type of input.



              If Input is in the form of String (i.e. "2 + 12 + 8") then the code will be -



              #include <stdio.h>

              int main()

              char str[] = "2 + 12 + 8";
              int sum=0;

              for (int i = 0; i < strlen(str); i++)
              if (str[i] >= '0' && str[i] <= '9')
              sum += str[i] - '0';

              printf("%d",sum);
              return 0;



              If Input is in the form of Array (i.e. [2, 12, 8]) then the code will be -



              #include <stdio.h>

              int main()

              int num[] = 2, 12, 8;
              int sum=0;
              int length = sizeof(num) / sizeof(num[0]);

              for (int i = 0; i < length; i++)
              while (num[i])
              sum += num[i] % 10;
              num[i] /= 10;


              printf("%d",sum);
              return 0;






              share|improve this answer



























                -1














                There is no direct formula for this task but you can perform this task in a separate function and use that function wherever it is needed.
                Here are two simple solutions which depend on the type of input.



                If Input is in the form of String (i.e. "2 + 12 + 8") then the code will be -



                #include <stdio.h>

                int main()

                char str[] = "2 + 12 + 8";
                int sum=0;

                for (int i = 0; i < strlen(str); i++)
                if (str[i] >= '0' && str[i] <= '9')
                sum += str[i] - '0';

                printf("%d",sum);
                return 0;



                If Input is in the form of Array (i.e. [2, 12, 8]) then the code will be -



                #include <stdio.h>

                int main()

                int num[] = 2, 12, 8;
                int sum=0;
                int length = sizeof(num) / sizeof(num[0]);

                for (int i = 0; i < length; i++)
                while (num[i])
                sum += num[i] % 10;
                num[i] /= 10;


                printf("%d",sum);
                return 0;






                share|improve this answer

























                  -1












                  -1








                  -1







                  There is no direct formula for this task but you can perform this task in a separate function and use that function wherever it is needed.
                  Here are two simple solutions which depend on the type of input.



                  If Input is in the form of String (i.e. "2 + 12 + 8") then the code will be -



                  #include <stdio.h>

                  int main()

                  char str[] = "2 + 12 + 8";
                  int sum=0;

                  for (int i = 0; i < strlen(str); i++)
                  if (str[i] >= '0' && str[i] <= '9')
                  sum += str[i] - '0';

                  printf("%d",sum);
                  return 0;



                  If Input is in the form of Array (i.e. [2, 12, 8]) then the code will be -



                  #include <stdio.h>

                  int main()

                  int num[] = 2, 12, 8;
                  int sum=0;
                  int length = sizeof(num) / sizeof(num[0]);

                  for (int i = 0; i < length; i++)
                  while (num[i])
                  sum += num[i] % 10;
                  num[i] /= 10;


                  printf("%d",sum);
                  return 0;






                  share|improve this answer













                  There is no direct formula for this task but you can perform this task in a separate function and use that function wherever it is needed.
                  Here are two simple solutions which depend on the type of input.



                  If Input is in the form of String (i.e. "2 + 12 + 8") then the code will be -



                  #include <stdio.h>

                  int main()

                  char str[] = "2 + 12 + 8";
                  int sum=0;

                  for (int i = 0; i < strlen(str); i++)
                  if (str[i] >= '0' && str[i] <= '9')
                  sum += str[i] - '0';

                  printf("%d",sum);
                  return 0;



                  If Input is in the form of Array (i.e. [2, 12, 8]) then the code will be -



                  #include <stdio.h>

                  int main()

                  int num[] = 2, 12, 8;
                  int sum=0;
                  int length = sizeof(num) / sizeof(num[0]);

                  for (int i = 0; i < length; i++)
                  while (num[i])
                  sum += num[i] % 10;
                  num[i] /= 10;


                  printf("%d",sum);
                  return 0;







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Mar 7 at 11:59









                  Arpit SethArpit Seth

                  192




                  192



























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