C-具有混合数字的整数数组

Question

I need to convert an integer array of size 4 into an int . I've seen solutions for int arrays that look like {1, 2, 3, 4} turn into 1234, and I've also seen ones where an array like {10, 20, 30} would turn into 102030. However, my problem is that I'll have an array that might look like {0, 6, 88, 54} and the solutions I previously mentioned only work on arrays with int s of the same type {eg all one digit or all two digit}.

What should I do to solve this?

My expected output from the {0, 6, 88, 54} array would be 68854.

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Examples An output with zeros in the middle should keep them, ie {6, 0, 0, 8} would be 6008 but {0, 6, 0, 0, 8} by default would still be 6008 in int form. I need this in an int but I wouldn't mind having a string intermediate.

Answer 1

You could do something like this:

int res = 0;
int nums[4] = {1, 4, 3, 2}

int i = 0, temp;
for (i = 0; i < 4; ++i) {
  temp = nums[i];
  do {
    res *= 10;
  } while ((temp /= 10) > 0);


  res += nums[i];
}

Answer 2

a neat solution would perhaps be to print to a string then convert the string back to an integer

char dummy[100];
int answer;
int input[4];

....

sprintf(dummy,"%d%d%d%d",input[0],input[1],input[2],input[3]);
answer=atoi(dummy);

the sprintf prints your integers into a string

the atoi converts the string into your integer, and it should be able to handle a 0 at the front.

full program

#include <stdio.h>
#include <stdlib.h>

int main()
{
  char dummy[100];
  int answer;
  int input[4]={3,4,0,345};

  sprintf(dummy,"%d%d%d%d",input[0],input[1],input[2],input[3]);
  answer=atoi(dummy);

  printf("%d\n",answer);

  return 0;
}

Answer 3

What about this solution?

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int arr[] = {0, 6, 88, 54};

    char buffer[1000] = { 0 };

    for(size_t i = 0; i < sizeof arr / sizeof *arr; ++i)
        sprintf(buffer, "%s%d", buffer, arr[i]);

    int val = strtol(buffer, NULL, 10);
    printf("%d\n", val);

    return 0;
}

int prints 608854 .