digits to english

[watshamacalit]

This program is supposed to output the digits of a number in English. so if the user inputs 100, the output should be "one, zero, zero".

i cant figure out whats wrong with my program...

#include <stdio.h>

void get_string( char buffer[], int size ) ;
char english_number( char string[] ) ;

void get_string( char buffer[], int size )
{
 	char character;
 	int j = 0;		

 	do									/* Get a character until newline or 		*/
 	{									/* we run out of characters.				*/
		character = getchar() ;
		buffer[j] = character ;
		++j;
	}
	while ( character != '\n' && j < size ) ;
	
	while ( character != '\n' )			/* Get rid of extra characters.				*/
		character = getchar() ;

	buffer[j - 1] = '\0' ;				/* Replace newline with the null-byte.		*/
}

void main( void )
{
	char line[81] ;
	int count ;
	char result, one, two, three, four, five, six, seven, eight, nine ;

	printf( "Enter a number  :  " ) ;
	get_string( line, 81 ) ;

	for( count = 0 ; line[count] != '\0' ; count++ )
	{
		english_number( line ) ;
		printf( "%s\n", line ) ;
	}

}

char english_number( char string[] )
{
	int count ;
	char one, two, three, four, five, six, seven, eight, nine ;

	for( count = 0 ; string[count] != '\0' ; count++ )
	{
		if( string[count] == '1' )
			return( 'one' ) ;
		else if ( string[count] == '2' )
			return( 'two' ) ;
		else if ( string[count] == '3' )
			return( 'three' ) ;
		else if ( string[count] == '4' )
			return( 'four' ) ;
		else if ( string[count] == '5' )
			return( 'five' ) ;
		else if ( string[count] == '6' )
			return( 'six' ) ;
		else if ( string[count] == '7' )
			return( 'seven' ) ;
		else if ( string[count] == '8' )
			return( 'eight' ) ;
		else if ( string[count] == '9' )
			return( 'nine' ) ;
	}

}

is there something wrong with the else if statements? or is there another way to do it? or am i completely lost?

[quzah]

char english_number( char string[] )

This should return a character pointer.

return( 'three' ) ;

Use double quotes for strings. ' ' is used for SINGLE CHARACTERS ONLY.

Quzah.

[Hammer]

>>void main( void )
is a big no no. It's int main(void) and return something at the end.

And please use code tags.

[watshamacalit]

ok...

i tried yer suggestions but it still doesnt work...

i keep getting this error

C:\Program Files\Microsoft Visual Studio\MyProjects\11\1101p04kt.cpp(54) : error C2440: 'return' : cannot convert from 'char [5]' to 'char'
This conversion requires a reinterpret_cast, a C-style cast or function-style cast

what does that mean?

[Hammer]

Change a couple of things:

>english_number( line ) ;
>printf( "%s\n", line ) ;
Replace these with
>printf ("%s\n", english_number( line ) );

And change the return type of english_number from char to char*.

[watshamacalit]

ok...that solved that error...

but now im getting this warning...

C:\Program Files\Microsoft Visual Studio\MyProjects\11\1101p04kt.cpp(16) : warning C4101: 'result' : unreferenced local variable

and it gives it for all my char variables in english_number

[Hammer]

It means you haven't actually used them, so there's no point in them being there.

Oh, I forgot, string is a reserved name (as is anything starting str followed by a letter), so don't use it

[ronin]

Whether it's proper coding or not is open to debate, but how about something like.

#include <stdio.h>

void printString(char []);

int main(void)
{
  char buffer[20];
	
  printf("Enter a number: ");
  fgets(buffer, 20, stdin);
  printString(buffer);
  return 0;
}

void printString(char source[])
{
  char *alphaValue[10] = { "Zero", "One", "Two", "Three",
                      "Four", "Five", "Six", "Seven",
		      "Eight", "Nine" };
  char digit;
  int i;
  
  for(i = 0; source[i] != '\0'; i++)
  {
    digit = source[i];
    switch(digit)
    {
      case '0': printf("%s ", alphaValue[0]);
	        break;
      case '1': printf("%s ", alphaValue[1]);
	        break;
      case '2': printf("%s ", alphaValue[2]);
	        break;
      case '3': printf("%s ", alphaValue[3]);
	        break;
      case '4': printf("%s ", alphaValue[4]);
	        break;
      case '5': printf("%s ", alphaValue[5]);
	        break;
      case '6': printf("%s ", alphaValue[6]);
	        break;
      case '7': printf("%s ", alphaValue[7]);
	        break;
      case '8': printf("%s ", alphaValue[8]);
	        break;
      case '9': printf("%s ", alphaValue[9]);
	        break;
      default:  break;
    };
  }
}

Your code above seems a little busy just to print out one, two, three....

[quzah]

x=isdigit(something);
printf("%s", x < 10 && x > -1 ? alphaValue[x] : "invalid" );

Quzah.

[ronin]

Doh... quzah got me again, but In my defense I'm merely a pseudo coder.... remember?.... :P

Hmm... I rather like that shortcut, yet another one added to my notebook. Thanks

[Hammer]

Originally posted by quzah
x=isdigit(something);
printf("%s", x < 10 && x > -1 ? alphaValue[x] : "invalid" );

Quzah.

eh? Are you talking about the C standard function isdigit() ? I presume not, as your example wouldn't work too well.

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

const char *alphaValue[] = 
{
    "Zero", "One", "Two", "Three", "Four",
    "Five", "Six", "Seven", "Eight", "Nine" 
};

void PrintString(char *source)
{
    for ( ; *source != '\0'; source++)
        if (isdigit(*source))
            printf ("%s ", alphaValue[*source - '0']);
        else printf("Invalid ");
    putchar('\n');
}

int main(void)
{
    PrintString("123Z45");
    return(0);
}
/*
Output
One Two Three Invalid Four Five
*/

[quzah]

Originally posted by Hammer
eh? Are you talking about the C standard function isdigit() ? I presume not, as your example wouldn't work too well.

No, that was me being side tracked while posting. I was looking at the post preceeding mine, and saw the reference to "digit" in there. For my excuse, I was actually on the phone at the time. I simply meant to validate the single character to ensure it was a number.

Convert x to an integer, x being a single character, and then apply it to that line.



Quzah.