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HughG

Basic string class problem

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I've decided to take a step back from game programming and brush up on some of my C++ basics. I picked up a book by Bjarne Stroustrup and am trying to do some of the exercises. Alright, so I'm trying to write a program that accepts a string (exercise calls for string and not a c style array of chars) and outputs the number of letters contained in the string. It does this by printing out the entire alphabet and the number of times a letter in the alphabet occured. Example: Input string: "abbbdsadjk" Output: "a: 2, b: 3, c: 0, d: 2, e: 0 ..... (entire alphabet) So this is what I came up with:
#include <string>
#include <iostream>
using namespace std;

void count(string*, struct Alphabet[]);

struct Alphabet{
	char letter;
	int number;
};


int main(){

	string blah = "xxadfjkljsduio";
	Alphabet alphabet[] = {{'a'}, {'b'}, {'c'}, {'d'}, {'e'}, {'f'}, {'g'}, {'h'}, {'i'}, {'j'}, 
		{'k'}, {'l'}, {'m'}, {'n'}, {'o'}, {'p'}, {'q'}, {'r'}, {'s'}, {'t'}, {'u'}, {'v'}, {'w'}, 
		{'x'}, {'y'}, {'z'}};
	
	count(&blah, alphabet);
	
	for(int i = 0; i <27; i++)		
		cout << alphabet.letter << " " << alphabet.number << "\n";
		
}

void count(string *s, Alphabet a[]){
	
	for(int i = 0; i < s->length(); i++){
		char temp = *s; //compiler doesn't like this line
		for(int j = 0; j < 26; j++){
			if(a[j].letter == temp)
				a[j].number++;
		}
        s++;
	}
	

}
Now I can't figure out how to compare the individual letters of the string to the individual characters inside the Alphabet struct. I looked up some methods in the string library and came across one called c.str() but I could not figure out how to make it work in this instance. Any ideas? Also, how could I modify the program to pass a pointer to an array of alphabet structs instead of passing the entire array of structures? TIA

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char temp = *s; //compiler doesn't like this line

you are trying to assign a string to a char here. You probably wanted to do
char temp = ( *s );

[edit] hmph, beaten
Also, I would suggest you use a reference instead of a pointer to pass the string into a function, like so: 'string *s' becomes 'const string &s', that way, you don't need to dereference the pointer above ( 'char temp = ( *s );' becomes 'char temp = s;' )

Regards,
jflanglois

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Since you're brushing up on your fundamental C++...

1. Use std::map.
#include <map>

std::map<char, int> Frequency;

...

// for each character in string
// if character is key in map, increment value
// else insert new key-value pair, with value as 1
As it stands, this does not represent zero-frequency items.

Using maps, vectors, strings and so forth also eliminates hard-coded numerical limits and other magic values in your code. Here's a complete example:

#include <algorithm>
#include <iostream>
#include <locale>
#include <map>
#include <string>

int main()
{
using namespace std;

char alphabet[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
map<char, size_t> Frequency;
for(int i = 0; i < strlen(alphabet); ++i)
Frequency.insert(pair&lt;char, size_t>(alphabet, 0));

string s;
cout << "Please enter a string of text: ";
getline(cin, s);

// first, convert the text to all uppercase for case-insensitive tally
transform(s.begin(), s.end(), s.begin(), toupper);

for(int i = 0; i < strlen(alphabet); ++i)
Frequency[alphabet] = count(s.begin(), s.end(), alphabet);

// print out the results
map<char, size_t>::iterator iter, stop = Frequency.end();
for(iter = Frequency.begin(); iter != stop; ++iter)
cout << iter->first << ": " << iter->second << endl;

return 0;
}
Obviously, there are inefficiencies in this code, but look how much more compact it is, and how much faster it is to write once you're familiar with all the constructs.

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jflanglois and ftn: Thanks, that is exactly what I was trying to do. I guess I assumed that the string pointer would point at the first letter in the string (like a char pointer to a char array).

Oluseyi: Beautiful! I really appreciate the advice. I have no experience with the STL at all and Stroustrup actually mentions the map class early in the book but fails to show any examples with it.

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