digits to english

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...

Code:

#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?

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.

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

And please use code tags. :)

digits to english

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?

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*.

digits to english

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

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 ;)

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

Code:

#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....

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

Quzah.

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 :)

Quote:

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.

Code:

#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 */

Quote:

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.