anagram program help

that one works todd, except that anagrams don't count things like whitespace and punctuation

Quote:

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Originally Posted by m37h0d

that one works todd, except that anagrams don't count things like whitespace and punctuation

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anagrams are word plays - they don't ever count anything!

Oh, you mean the program. Yes, well, unfortunately, it counts everything. Therefore, use what you want and ignore the rest.

edit: i got it to work using this code .. now does anyone know if i should just count the letter .. or can an anagram use numbers as well?

Code:

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#include <iostream>
#include <cctype>
#include <cstring>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

using namespace std;


struct characters {
        int letters[256] ;
} ;

void map(const char * s, struct characters * c) {
        memset( c, 0, sizeof(struct characters) ) ;       
        while(*s) {
                if (*s != ' ') {
                        c->letters[(int) tolower(*s)]++ ;
                }
                s++ ;
        }
}

int compare( const struct characters * c1, const struct characters * c2  ) {
        return !memcmp( c1, c2, sizeof(struct characters)) ;
}


int main() {
        struct characters c1 , c2 ;

        char s1[500];
        char s2[500];
        char mystring1[500];
        char mystring2[500];

        cout << "enter first string" << endl;
        cin.getline(mystring1, 500, '\n');
 int a=0;
        for (int i=0; mystring1[i] !='\0'; i++){
                if (isalpha(mystring1[i])){

                        s1[a]=mystring1[i];
                        a++;
                }
        }
s1[a]='\0';
        cout << s1 << endl;



        cout << "enter second string" << endl;
        cin.getline(mystring2, 500, '\n');

        int b=0;
        for (int i=0; mystring2[i] !='\0'; i++){
                if (isalpha(mystring2[i])){

                        s2[b]=mystring2[i];
                        b++;
                }
        }
s2[b]='\0';
        cout << s2 << endl;


       
        map(s1, &c1) ;
        map(s2, &c2) ;
        printf("The strings &#37;s Anagrams\n", (compare( &c1, &c2 ) ? "are" : "are not" ) ) ;

        return 0 ;




}

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Well, since you need to be writing in C++, the only thing you should be taking out of my sample is the concept of how to do it.

Todd

well i don't fully understand how your function works .. for instance .. i do not know what struct or memset are

if you were able to explain the function a little more i would understand it .. but like i said what i have now works

Code:

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mem.h, string.h

Category

Memory and String Manipulation Routines, Inline Routines

Prototype

void *memset(void *s, int c, size_t n);

void *_wmemset(void *s, int c, size_t n);

Description

Sets n bytes of a block of memory to byte c.

memset sets the first n bytes of the array s to the character c.

Return Value

memset returns s.

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Code:

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Category

Type specifiers

Syntax

struct [<struct type name>] {

  [<type> <variable-name[, variable-name, ...]>] ;
    .
    .
    .

} [<structure variables>] ;

Description

Use a struct to group variables into a single record.

<struct type name>        An optional tag name that refers to the structure type.
<structure variables>        The data definitions, also optional.

Though both <struct type name> and <structure variables> are optional, one of the two must appear.

You define elements in the record by naming a <type>, followed by one or more <variable-name> (separated by commas).

Separate different variable types by a semicolon.

To access elements in a structure, use a record selector (.) or ->.

To declare additional variables of the same type, use the keyword struct followed by the <struct type name>, followed by the variable names. In C++ the keyword struct can be omitted.

Note: C++Builder allows the use of anonymous struct embedded within another structure.

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learning to read manual-ese is by far the most valuable skill you will ever learn in programming.