How do I change intergers in a string to their word equivalent (ie 1=one, 2=two)

Code:

---

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

int main(int argc, char *argv[])
{
    int n1;
    int i=0;
    char str1[100];
    char str2[100];
   
    printf("Please enter 1st number: ");
    scanf("%s", str1);
   
    printf("\nPlease enter 2nd number: ");
    scanf("%s", str2);
   
    printf("\nConcatenation of 1st & 2nd: %s\n", strcat(str1, str2));
    getchar();
   
    n1 = atoi( strcat(str1, str2) );
    printf("\nOnly Digits of the Concatenation: %d\n", n1);
   
   
    system("PAUSE");
    return 0;
}

---

This program so far asks for 1st string and 2nd string, then adds them together, then removes any letters, the next part I need to change all the digits its found to their word equivalent (ie 1=one, 2=two)

There's no API that does this conversion for you, so you'll have to have some kind of data structure that maps numbers to names. Obviously you don't want to do this for every single number, just the ones that have a single word that represents them, then combine the words based on whatever value you have.

Code:

---

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


void foo(int value);

int main(int argc, char *argv[])
{
    char line[100];
    int i, n1, n2, num;

   
    char str1[100];
    char str2[100];
   
    printf("Please enter 1st number: ");
    scanf("%s", str1);
   
    printf("\nPlease enter 2nd number: ");
    scanf("%s", str2);
   
    printf("\nConcatenation of 1st & 2nd: %s\n", strcat(str1, str2));
    getchar();
   
    n1 = atoi( strcat(str1, str2) );
    printf("\nOnly Digits of the Concatenation: %d\n", n1);
   
    foo(n1);
    putchar('\n');

   
   
    system("PAUSE");
    return 0;
}

void foo(int value)
{
  static const char *digit[] =
  {
      "zero", "one", "two", "three", "four",
      "five", "six", "seven", "eight", "nine",
  };
  if ( value > 0 )
  {
      foo(value / 10);
      printf("%s ", digit [ value % 10 ] );
  }
}

---

I almost have it, this doesnt covert fully if theres a word/letter in the n1 =/

Code:

---

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

void foo(int value);


int main(int argc, char *argv[])
{
    int n1, n2, i=0;
    char str1[100];
    char str2[100];
   
    printf("Please enter 1st number: ");
    scanf("%s", str1);
   
    printf("\nPlease enter 2nd number: ");
    scanf("%s", str2);
   
    strcat(str1, str2);
    printf("\nConcatenation of 1st & 2nd: %s\n", str1);
   
    n1 = atoi( str1 );
    printf("\nOnly Digits of the Concatenation: %d\n", n1);
   
    n2 = n1;
    foo(n2);
   
   

    system("PAUSE");
    return 0;
}

void foo(int value)
{
  static const char *digit[] =
  {
      "zero", "one", "two", "three", "four",
      "five", "six", "seven", "eight", "nine",
  };
  if ( value > 0 )
  {
      foo(value / 10);
      printf("%s ", digit [ value % 10 ] );
  }
}

---

^works if no letters are entered, but I want it to take out the letters and concat

ie

1st string: 123
2nd string: 4dog56
concat: 123456
convert: one two three four five six

atoi stops the integer conversion when it encounters a character. You'll need to create a new string that only contains numbers and work with that.

Actually, a better way to handle this is to just take each character one at a time, and if it's a digit print the name. You can convert a character representing a digit by subtracting '0' (i.e. the ascii value for the character 0). That way, you don't have to worry about running over INT_MAX.