C语言-添加产品的数字（即不是产品本身）

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?

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 = %u\n", sum(a, sizeof(a)/sizeof(*a)));
}

Compilation and execution :

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

Answer 2

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("%d\n", 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).

Answer 3

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 (ie "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 (ie [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;
}