C - converting words into numbers

selent · #1 12-Dec-2005, 08:27

Hi,
I'm not very experienced in C and i'm trying to write the code for a predictive text program like T9.

The program should read in words from words.txt (it's a document with 25143 words), which it's doing correctly i think, and it's meant to return numbers.
Example:
'and'
'for'
should return as: 263
367
At the moment, i'm getting a huge list of numbers which i think are random. The problem is: I don't know what's wrong with the code. It should convert the words one by one into the numbers they represent. I will really appriciate it if someone could tell me what's wrong. I've highlited the 'wrong' parts red.

My code is the following:

CPP / C++ / C Code:

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

char* words[25143];

char LowerCaseValue(char letter)
{
   if ((letter >= 'A') && (letter <= 'Z'))
      return (letter + 32);
   else
      return letter;
}

unsigned let2int(char letter)
{
   unsigned result;

   if ((letter >= 'a') && (letter <= 'c'))
   {
      result=2;
   }
   else if ((letter >= 'd') && (letter <= 'f'))
   {
      result=3;
   }
   else if ((letter >= 'g') && (letter <= 'i'))
   {
      result=4;
   }
   else if ((letter >= 'j') && (letter <= 'l'))
   {
      result=5;
   }
   else if ((letter >= 'm') && (letter <= 'o'))
   {
      result=6;
   }
   else if ((letter >= 'p') && (letter <= 's'))
   {
      result=7;
   }
   else if ((letter >= 't') && (letter <= 'v'))
   {
      result=8;
   }
   else if ((letter >= 'w') && (letter <= 'z'))
   {
      result=9;
   }
   return result;
}

[color="red"]void word2numeric(char* word, unsigned numeric[30])
{
   register int i;
   int count=0;

   
for (i=0;i<(int)strlen(word);i++)
   {
      numeric[count]=let2int(LowerCaseValue(word[i]));
      count++;
   }
}[/color]int main(void)

{	
	FILE *data_in;
	unsigned numeric[30];
	char * word;
	int i;
	int count=0;
	char words[25143];

	data_in = fopen("words.txt","r");

	if (data_in == NULL) printf("Cannot open file\n");
	else
	{	
		while (feof(data_in) == 0)
		{
		fscanf(data_in, "%s", words);
		printf("selen");
			
			[color="Red"]word2numeric(word,numeric);
			for (i=0;i<25143;i++)
			{
			printf("%2d",numeric[i]);
			}[/color]		}
	}	
	fclose(data_in);   	
return 0;
}

Blake · #2 12-Dec-2005, 09:09

This will convert a char to its ascii code:

CPP / C++ / C Code:


inline unsigned short char2ascii(char C)
{
   return *((unsigned short*)&C);
}

selent · #3 12-Dec-2005, 10:41

I'd like to convert all words into numbers which appear on a mobile phone. i.e
a,b,c = 2
d,e,f = 3
g,h,i = 4 and so on..

WaltP · #4 12-Dec-2005, 11:59

Define an array of 26 integers corresponding to the letter/number combinations:

CPP / C++ / C Code:

int num[26] = 
   {2,2,2,    // abc 
    3,3,3,    // def
    4,4,4,    // ghi
    5,5,5,    // jkl
    6,6,6,    // mno
    7,7,7,7,  // pqrs
    8,8,8,    // tuv
    9,9,9     // wxyz
   }

In code:
1) get a single character, say chr
2) subtract 65 (A) to move it into the numbers 0-25 (A=65 => 0, B=66 => 1)
3) If chr < 0, it's invalid
4) if chr >= 26 subtract 6 because it's probably lower case
5) If chr < 0 or >= 26, it's invalid
6) num[chr] is your integer value

there is a mathmatical way to do this but it's more difficult.

Blake · #5 12-Dec-2005, 12:14

CPP / C++ / C Code:


unsigned short char2num(char C)
{
  string alphabet = "abccdeffghiijkllmnoopqrstuvvwxyz";
  return alphabet.find(C) / 4 + 2;
}

This works if you give it a valid input. The repeated characters are deliberate.

WaltP · #6 12-Dec-2005, 12:21

Quote:

Originally Posted by Blake

CPP / C++ / C Code:


unsigned short char2num(char C)
{
  string alphabet = "abccdeffghiijkllmnoopqrstuvvwxyz";
  return alphabet.find(C) / 4 + 2;
}

Convoluted and obfuscated. Let's try to keep it simple for the new programmers. This code also will not work with upper case letters. And what happens if you pass in the character '%' or '4' -- any non-letter?

Quote:

Originally Posted by Blake after an edit

This works if you give it a valid input. The repeated characters are deliberate.

And don't modify your post after someone comments on it to make their comment invalid. You don't really want to make people look like a dunderhead, do you?

Blake · #7 12-Dec-2005, 12:26

You left off d*** clever. (=

Often the cleverest solutions are hardest to understand.

As I said you would need to check for invalid characters before calling the function, but that would apply to any solution to this problem, so I don't see what the big deal there is.

Quote:

And don't modify your post after someone comments on it to make their comment invalid. You don't really want to make people look like a dunderhead, do you?

No one had responded yet when I made the modification.

WaltP · #8 12-Dec-2005, 12:34

Quote:

Originally Posted by Blake

No one had responded yet when I made the modification.

You must have been editing the post while I was commenting then... It happens...

Blake · #9 12-Dec-2005, 12:37

Well, sorry about the confusion.

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