ERROR! cannot convert parameter...

my compiler keeps giving me this error...

C:\Program Files\Microsoft Visual Studio\MyProjects\11\1101x01kt.cpp(25) : error C2664: 'english_number' : cannot convert parameter 1 from 'char [81]' to 'char'
This conversion requires a reinterpret_cast, a C-style cast or function-style cast

what does that mean?

this program is supposed to write the numbers in english( the way we really say them )...from 0-100...i've done it all but i keep getting this error so i've only posted the first half of my code cuz its so long...

heres my code...

Code:

#include <stdio.h> char* english_number( char string ) ; void get_string( char buffer[], int size ) ; const int LENGTH = 81 ; void main( void ) { char line[ LENGTH ] ; int count ; printf( "Enter a number : " ) ; get_string( line, LENGTH ) ; printf( "English : " ) ; for( count = 0 ; line[count] != '\0' ; count++ ) /*error here -->*/ printf( "%s ", english_number( line ) ) ; } char* english_number( char string ) { if ( string >= '0' && string <= '10' ) { if ( string == '0' ) return ( "zero" ) ; else if ( string == '1' ) return( "one" ) ; else if ( string == '2' ) return( "two" ) ; else if ( string == '3' ) return( "three" ) ; else if ( string == '4' ) return( "four" ) ; else if ( string == '5' ) return( "five" ) ; else if ( string == '6' ) return( "six" ) ; else if ( string == '7' ) return( "seven" ) ; else if ( string == '8' ) return( "eight" ) ; else if ( string == '9' ) return( "nine" ) ; else if ( string == '10' ) return( "ten" ) ; } } 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. */ }

>char* english_number( char string )
You pass line to this function, but line is an array of char, change the function tag to

char* english_number( char *string )

You'll also have to remodel the function since it uses the relational operators which don't work on strings.

>void main( void )
main returns an int, anything else is wrong.

-Prelude

thanks for your help

but why is there a black dot on my posts?

Code:

#include <stdio.h> char english_number( char *string ) ; void get_string( char buffer[], int size ) ; const int LENGTH = 81 ; int main( void ) { char line[ LENGTH ] ; int count ; printf( "Enter a number : " ) ; get_string( line, LENGTH ) ; printf( "English : " ) ; for( count = 0 ; line[count] != '\0' ; count++ ) /*error here -->*/ printf( "%s ", english_number( line ) ) ; } char english_number( char *string ) { if ( string >= '0' && string <= '10' )//have to change way they are compared { if ( (strcmpi(string, '0'))!= 0 ) return ( "zero" ) ;//cannot return a string like this else if ((strcmpi(string, '1'))!= 0 ) return( "one" ) ;//ditto else if ((strcmpi(string, '2'))!= 0 ) return( "two" ) ;//ditto else if ((strcmpi(string, '3'))!= 0 ) return( "three" ) ;//ditto else if ((strcmpi(string, '4'))!= 0 ) return( "four" ) ;//ditto else if ( (strcmpi(string, '5'))!= 0 ) return( "five" ) ;//so forth... else if ( (strcmpi(string, '6'))!= 0 ) return( "six" ) ; else if ( (strcmpi(string, '7'))!= 0 ) return( "seven" ) ; else if ( (strcmpi(string, '8'))!= 0 ) return( "eight" ) ; else if ( (strcmpi(string, '9'))!= 0 ) return( "nine" ) ; else if ((strcmpi(string, "10"))!= 0 ) return( "ten" ) ; } } 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. */ }

Not everything is fixed to how it should be and I purposely did somethings the wrong way and just didn't do others because what fun is it for you to get the code completed and done for you.:D

Code:

#include <stdio.h> char english_number( char *string ) ; void get_string( char buffer[], int size ) ; const int LENGTH = 81 ; int main( void ) { char line[ LENGTH ] ; int count ; printf( "Enter a number : " ) ; get_string( line, LENGTH ) ; printf( "English : " ) ; for( count = 0 ; line[count] != '\0' ; count++ ) /*error here -->*/ printf( "%s ", english_number( line ) ) ; } char english_number( char *string ) { if ( string >= '0' && string <= '10' )//have to change way they are compared { if ( (strcmpi(string, '0'))!= 0 ) return ( "zero" ) ;//cannot return a string like this else if ((strcmpi(string, '1'))!= 0 ) return( "one" ) ;//ditto else if ((strcmpi(string, '2'))!= 0 ) return( "two" ) ;//ditto else if ((strcmpi(string, '3'))!= 0 ) return( "three" ) ;//ditto else if ((strcmpi(string, '4'))!= 0 ) return( "four" ) ;//ditto else if ( (strcmpi(string, '5'))!= 0 ) return( "five" ) ;//so forth... else if ( (strcmpi(string, '6'))!= 0 ) return( "six" ) ; else if ( (strcmpi(string, '7'))!= 0 ) return( "seven" ) ; else if ( (strcmpi(string, '8'))!= 0 ) return( "eight" ) ; else if ( (strcmpi(string, '9'))!= 0 ) return( "nine" ) ; else if ((strcmpi(string, "10"))!= 0 ) return( "ten" ) ; } } 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. */ }

I did not fix everything for you and purposely made somethings the wrong way because it wouldn't be to fun for you if I fixed everything :D

>but why is there a black dot on my posts?
The black dot is there to let you know that you've posted to a thread.

>return ( "zero" ) ;//cannot return a string like this
Why not?

>strcmpi(string, '0'))!= 0
What purpose does using a nonstandard case insensative string comparison serve when case is irrelevant. The last time I checked, digits don't have upper and lower case.

Wolfpack if you're going to post code, make sure it's at least correct. If you put errors in that code on purpose, please think about why you're here bfore posting again. Also, the // comment is only valid in C99 or C++, most likely you don't have a compiler for the former and the latter is off topic for this forum.

Code:

char *english_number( char *string ) { if ( strcmp ( string, "0" ) == 0 ) return ( "zero" ) ; else if ( strcmp ( string, "1" ) == 0 ) return( "one" ) ; else if ( strcmp ( string, "2" ) == 0 ) return( "two" ) ; else if ( strcmp ( string, "3" ) == 0 ) return( "three" ) ; else if ( strcmp ( string, "4" ) == 0 ) return( "four" ) ; else if ( strcmp ( string, "5" ) == 0 ) return( "five" ) ; else if ( strcmp ( string, "6" ) == 0 ) return( "six" ) ; else if ( strcmp ( string, "7" ) == 0 ) return( "seven" ) ; else if ( strcmp ( string, "8" ) == 0 ) return( "eight" ) ; else if ( strcmp ( string, "9" ) == 0 ) return( "nine" ) ; else if ( strcmp ( string, "10" ) == 0 ) return( "ten" ) ; return ""; /* No valid comparisoon */ }

-Prelude

I cringe at pointers as they are not my friend; however, if a real programmer would be so kind as to correct my usage, then this crud *might* work as a starting point though I'm not sure how you'd tell the difference between 10 and 1 0 in such a string of numbers now that I re-read your topic..

Code:

#include <stdio.h> #include <malloc.h> #include <stdlib.h> char *toEnglish(char []); // unbounded arrays are cool int main(void) { char number[80]; char *returnedptr; int i, length; printf("Enter a number: "); fgets(number, 80, stdin); for(i=0; number[i] != '\0'; i++) length = i; // cheap knock off for strlen(string) number[length] = '\0'; // get rid of '\n' if((returnedptr = (char *) malloc (256)) == NULL) { printf("Unable to allocate memory for conversion."); return -1; } returnedptr = toEnglish(number); printf("%s", returnedptr); return 0; } char *toEnglish(char conversion[]) { char digit; int count, j, k; const char *EnglishArray[11] = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "invalid" }; static char strptr[256]; k=0; for(count = 0; conversion[count] != '\0'; count++) { digit = conversion[count]; switch(digit) { case '0': for(j = 0; EnglishArray[0][j] != '\0'; j++) strptr[k++] = EnglishArray[0][j]; strptr[k++] = ' '; break; case '1': for(j = 0; EnglishArray[1][j] != '\0'; j++) strptr[k++] = EnglishArray[1][j]; strptr[k++] = ' '; break; case '2': for(j = 0; EnglishArray[2][j] != '\0'; j++) strptr[k++] = EnglishArray[2][j]; strptr[k++] = ' '; break; case '3': for(j = 0; EnglishArray[3][j] != '\0'; j++) strptr[k++] = EnglishArray[3][j]; strptr[k++] = ' '; break; case '4': for(j = 0; EnglishArray[4][j] != '\0'; j++) strptr[k++] = EnglishArray[4][j]; strptr[k++] = ' '; break; case '5': for(j = 0; EnglishArray[5][j] != '\0'; j++) strptr[k++] = EnglishArray[5][j]; strptr[k++] = ' '; break; case '6': for(j = 0; EnglishArray[6][j] != '\0'; j++) strptr[k++] = EnglishArray[6][j]; strptr[k++] = ' '; break; case '7': for(j = 0; EnglishArray[7][j] != '\0'; j++) strptr[k++] = EnglishArray[7][j]; strptr[k++] = ' '; break; case '8': for(j = 0; EnglishArray[8][j] != '\0'; j++) strptr[k++] = EnglishArray[8][j]; strptr[k++] = ' '; break; case '9': for(j = 0; EnglishArray[9][j] != '\0'; j++) strptr[k++] = EnglishArray[9][j]; strptr[k++] = ' '; break; default : for(j = 0; EnglishArray[10][j] != '\0'; j++) strptr[k++] = EnglishArray[10][j]; strptr[k++] = ' '; break; }; // end switch } // end for loop strptr[k] = '\0'; return strptr; }

I rather like arrays though. :P

If you're going to fill the screen with code you might as well cover more than 0-10:

Code:

#include <stdio.h> #include <string.h> char *wordbuild ( char *n, int ncomm ) { int i, nleft, len; char *p = NULL; static char ret[1024]; /* Word lists */ static char *ones[] = {"one ","two ","three ","four ", "five ","six ","seven ","eight ","nine "}; static char *tens[] = {"ten ","eleven ","twelve ","thirteen ", "fourteen ","fifteen ","sixteen ","seventeen ","eighteen ","nineteen "}; static char *twenties[] = {"","twenty ","thirty ","forty ", "fifty ","sixty ","seventy ","eighty ","ninety "}; static char *hundreds[] = { "hundred ","thousand ","million "}; memset ( ret, '\0', 1024 ); len = strlen ( n ); for ( i = 0; i < len; i++ ) { if ( ( p = strchr ( ( n[i] == ',' ) ? &n[++i] : &n[i], ',' ) ) == NULL ) p = &n[strlen ( n )]; if ( n[i] == '0' ) continue; if ( ( nleft = ( p - &n[i] ) ) != 0 ) { if ( nleft == 3 ) { strcat ( ret, ones[n[i] - '0' - 1] ); strcat ( ret, hundreds[0] ); } else if ( nleft == 2 ) { if ( n[i] == '1' ) { strcat ( ret, tens[n[++i] - '0'] ); nleft--; } else strcat ( ret, twenties[n[i] - '0' - 1] ); } else strcat ( ret, ones[n[i] - '0' - 1] ); } if ( nleft == 1 && ncomm != 0 ) strcat ( ret, hundreds[ncomm--] ); } return ret; } char *commaize ( unsigned long n, int *ncomm ) { static char ret[30]; int i = 0; char *p = &ret[sizeof(ret)-1]; *p = '\0'; *ncomm = 0; do { if ( i % 3 == 0 && i != 0 ) { *--p = ','; ++*ncomm; } *--p = (char)( '0' + n % 10 ); n /= 10; i++; } while ( n != 0 ); return p; } char *stow ( int n, char *buf ) { int i; char *p, *ret; p = commaize ( n, &i ); ret = wordbuild ( p, i ); if ( buf != NULL ) { strcpy ( buf, ret ); return buf; } return ret; } int main ( void ) { char buf[1024]; printf ( "%s\n", stow ( 123456789, NULL ) ); printf ( "%s\n", stow ( 12, NULL ) ); stow ( 23, buf ); printf ( "%s\n", buf ); return 0; }

:D

-Prelude

Quote:

Originally posted by Prelude
If you're going to fill the screen with code you might as well cover more than 0-10:
-Prelude

lol.... well it isn't my program besides, I only picked up a C book last weekend. :P

A million... that's funny.;)

Now that's what I call coding..... let me copy that paste to see what it does...

[output]one hundred twenty three million four hundred fifty six thousand seven hundred eighty nine
twelve
twenty three [/output]

Could use a comma or something. :D

Thanks Prelude.... that code is interesting.

Please Prelude take the time to look at the last line of the post that I made.

Quote:

Not everything is fixed to how it should be and I purposely did somethings the wrong way and just didn't do others because what fun is it for you to get the code completed and done for you.

I see no way that he could learn from getting code back that is perfect. I learn on the basis of getting a push from someone else and going back and fixing it myself. Part of programming is to know what is going on and what ways are better than others and the only way that you gain this is through experience which is gained by going back and fixing itself.

>Please Prelude take the time to look at the last line of the post that I made.
I did not miss that part of your post, but my comments still stand. Someone comes onto these forums looking for an answer to a question that supposedly they've been working on and couldn't figure out. The last thing they want to hear is "I'll make a few inaccurate comments about your code and leave fixing the errors up to you, enjoy!". I find such posts incredibly arrogant and rude.

>I see no way that he could learn from getting code back that is perfect.
Perhaps correct and working examples that are explained thoroughly as well as what was wrong with their code to begin with? I see no way that he could learn from being given back the same errors he couldn't fix so that he gets discouraged and quits.

>I learn on the basis of getting a push from someone else and going back and fixing it myself.
Not everyone is like you.

-Prelude

*bows his head to superiority*

Well I didn't have time to go a bit fancier. I just edited some stuff and I haven't tested it so use at your own discretion but besides why would you be lookin at mine when you got Prelude's up there :D

Code:

#include <stdio.h> char english_number( char *string ) ; void get_string( char buffer[], int size ) ; const int LENGTH = 81 ; char *NUMB[11]={"zero","one","two","three", "four","five","six","seven", "eight","nine","ten"}; int main( void ) { char line[ LENGTH ] ; int count ; printf( "Enter a number : " ) ; get_string( line, LENGTH ) ; printf( "English : " ) ; for( count = 0 ; line[count] != '\0' ; count++ ) /*error here -->*/ printf( "%s ", english_number( line ) ) ; } char english_number( char *string ) { if ( (strcmp(string, "0"))== 0 ) return ( NUMB[0] ) ; else if ((strcmp(string, "1"))== 0 ) return( NUMB[1] ) ; else if ((strcmp(string, "2"))== 0 ) return( NUMB[2] ) ; else if ((strcmp(string, "3"))== 0 ) return( NUMB[3] ) ; else if ((strcmp(string, "4"))== 0 ) return( NUMB[4] ) ; else if ( (strcmp(string, "5"))== 0 ) return( NUMB[5] ) ; else if ( (strcmp(string, "6"))== 0 ) return( NUMB[6] ) ; else if ( (strcmp(string, "7"))== 0 ) return( NUMB[7] ) ; else if ( (strcmp(string, "8"))== 0 ) return( NUMB[8] ) ; else if ( (strcmp(string, "9"))== 0 ) return( NUMB[9] ) ; else if ((strcmp(string, "10"))== 0 ) return( NUMB[10] ) ; } 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. */ }

thanks for your help

this will help a lot, thanks guys

Prelude, in your code above could you explain the role of memset?

Could you also explain the role of *ncomm in commarize and ncomm in word build.

The commarize code is a little confusing, once I understand that I think I may understand how to do this problem!

>Prelude, in your code above could you explain the role of memset?
It clears out the contents from previous calls to the function.

>Could you also explain the role of *ncomm in commarize and ncomm in word build.
It holds the number of commas. This is how I determined where to print certain qualifiers such as hundred, thousand and million. At the time I wrote this I determined that that approach would be simpler than counting digits.

On a side note, please don't reply to such old threads. If you have a question to me personally then please contact me via email or preferably by private message.