converting a number into words

[Guest Adriano]

I'm studying C programming with "Practical C programming 3rd ed" by Steve Oualline (O'reilly). It's a very good book, and provides a series of exercises at the end of each chapter to test your knowledge so far.

 

I'm now stumbling at chapter 8. The exercise (8-6 if you happen to have the book) asks a program to change numbers into words, e.g. 895 is "eight nine five". I can't think of an algorithm to traverse the numbers and convert them. I could take them from the input as strings (fgets...) but I run into trouble trying to compare them... I also tried dividing by ten to get the tenths, hundredths, etc.. but haven't reached a good algorithm so far.

 

Because of the position in the book, I am supposed to use only control statements, but not pointers, functions, structs or anything too complex. While I can use some library functions (maybe atoi() or something), I guess the purpose is for me to write my own.

 

Can anyone give me a pointer to something?

[tyme]

does C have the char type (I'm a C++ person)? you could use a char array...

char nums[3];

then you can seperate them out using if/else if or case, and referring to the specific one in the array:

nums[X]

X being 0 - 2 (since arrays start at 0)

[Guest Adriano]

Yes, C has char arrays. Actually, strings in C are only char arrays ended in

'\0'

. But this presupposes a fixed length for the numbers, doesn't it (I'm a perfectionist)? Or else a long string of zeros until the numbers, or something... I guess I have to reread the arrays section.

[Guest Adriano]

Solved, with help from a friend. It'd possibly be simpler with an array, but anyway here's the code:

 

/*86numwords.c - Exercise 8-6 - Converts numbers into words */
#include <stdio.h>
#include <stdlib.h>

char line[80]; /* the line buffer. Stores the numbers as chars */
int i; /* just a for loop counter */
int main(void){
 printf("86numwords\nConverts numbers into words\n");
 while(1) {
     printf("Please write a number (or anything else to quit): ");
     fgets(line, sizeof(line),stdin);
     line[strlen(line)-1] = '\0'; /* getting rid of the \n at the end of the string */
   
     for (i = 0; i < strlen(line); ++i) {/* this for traverses through the character array that stored the number */
         switch(line[i]) { /* ... And this switch comes up with each digit as word. */
             case '0': {
                 printf("zero ");
                 break;
             }
             case '1': {
                 printf("one ");
                 break;
             }
             case '2': {
                 printf("two ");
                 break;
             }
             case '3': {
                 printf("three ");
                 break;
             }
             case '4': {
                 printf("four ");
                 break;
             }
             case '5': {
                 printf("five ");
                 break;
             }
             case '6': {
                 printf("six ");
                 break;
             }
             case '7': {
                 printf("seven ");
                 break;
             }
             case '8': {
                 printf("eight ");
                 break;
             }
             case '9': {
                 printf("nine ");
                 break;
             }
             default: { /* A quick-n-dirty way to exit*/
           printf("That's not a valid number\n");
           return(8);
             }
         }    
     }
     printf("\n"); /* we want a carriage return after the number completes. */
     system("PAUSE"); /* Good in Windows / Dev-cpp to keep the console window from closing automatically. Delete it in Linux. */
     }    
 return 0;
}

[Guest Adriano]

As promised, here's the same program done with an array. Certainly more compact, though I don't know which is more efficient.

/*86numwords.c - Exercise 8-6 - Converts numbers into words */
#include <stdio.h>
#include <stdlib.h>

char line[80]; /* the line buffer. Stores the numbers as chars */
int i; /* just a for loop counter */
char numbers[10][6] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",}; /*This two dimensional array stores 10 strings of 6 chars maximum length (remember that one char space is used by the \0 termination char) */

int main(void) {
 printf("86numwords\nConverts numbers into words\n");
 while(1) { /* This program doesn't stop. Close it with Control + C */
     printf("Please write a number: ");
     fgets(line, sizeof(line),stdin);
     line[strlen(line)-1] = '\0'; /* getting rid of the \n at the end of the string */
   
     for (i = 0; i < strlen(line); ++i) { 
         /* this for loop traverses through the character array that stored the number */

         int goodline = line[i] - '0'; /* This transforms all the input from chars to ints. */
         printf("%s ", numbers[goodline]);
     }
     printf("\n"); /* we want a carriage return after the number completes. */
     system("PAUSE"); /* Good in Windows / Dev-cpp to keep the console window from closing automatically. Delete it in Linux. */
     }    
 return 0;
}