CODEWITHSUNDEEP
c++string
johny tran
johny tran

Reputation: 31

how can i count how many consonants are in a string so far i have this

int countConsonant(string str, int consonant)
{
    int length = str.length();
     consonant = 0;

        for (int i = 0; i < length; i++)
        {
            if(str[i] == 'b'&& str[i] == 'c' && str[i] == 'd'&& str[i] == 'f'
                && str[i] == 'g'&& str[i] == 'h'&& str[i] == 'j' && str[i] == 'k'&& str[i] == 'l'&& str[i] == 'm'
                && str[i] == 'n'&& str[i] == 'p'&& str[i] == 'q'&& str[i] == 'r'&& str[i] == 's'&& str[i] == 't'
                && str[i] == 'v'&& str[i] == 'w'&& str[i] == 'x'&& str[i] == 'y'&& str[i] == 'z')

                consonant = consonant + 1;
        }
        return consonant;
}

Upvotes: 0

Views: 6804

Answers (8)

Muhammad Abdullah
Muhammad Abdullah

Reputation: 1

You can use the following to count how many consonants are in a string:

int main()
{
    char a[20];
    int consonants; 
    gets(a);
    cout<<a<<endl;
    if(a[i]>=0 && a[i]<=9)  //  consonants
    {
        consonants++;
    }
    cout<<"Total Consonats are: "<<consonants;
}

Upvotes: 0

Tony Delroy
Tony Delroy

Reputation: 106196

Way overkill for this homework, but just for reference - an object-oriented approach:

struct Consonants
{
    bool x_[256];

    Consonants()
    {
        // constructor sets x_[i] to true or false to indicate
        // whether character with ASCII number i is a consonant...

        for (int i = 0; i < 256; ++i)
            x_ = false;

        // note: ASCIIZ string finishes with NUL, so *p becomes false
        //       and for loop exits
        for (const char* p = "bcdfghjklmnpqrstvwxyz"
                             "BCDFGHJKLMNPQRSTVWXYZ"; *p; ++p)
            x_[*p] = true;
    };

    int in(const std::string& str) const
    {
        int result = 0;
        for (int i = 0; i < str.size(); ++i)
            if (x_[(unsigned char)str[i]])
                ++result;
        return result;
    }
};

Usage:

Consonants consonants;  // create an "utility" object once

int c1 = consonants.in("hello world");   // use as often as desired
int c2 = consonants.in("goodbye cruel world");

Upvotes: 0

Tony Delroy
Tony Delroy

Reputation: 106196

You were close, but should check with logical-or ||, not logical-and && (str[i] simply can't be equal to two different things).

The C++03 Standard allows you to use the keywords and and or - and not, xor etc - instead of these cryptic (for new programmers) symbols, but this hasn't caught on widely - perhaps because Microsoft's compiler doesn't default to being Standard-compliant in this regard - presumably to avoid breaking existing client code that has variables, functions, types etc. with these names. So, for portability and simplicity, many libraries and textbooks avoid these keywords too.

Another approach that might be a little more concise is to use isalpha() from <cctype> then check it's not a vowel. Faster approaches tend to use arrays from character value to bool, but beware indexing outside the array due to signed character values or >=128 bit non-ASCII values. If there's also uppercase/lowercase to consider - you may want to use tolower() on your character before testing it (i.e. char c = tolower(str[i])); if (c == '...).

Other notes: your function should:

  • accept its std::string argument by const reference (i.e. const std::string& str) to avoid unnecessary and time-consuming copying of the value from the calling context into a separate variable local to this function. The copying doesn't do any real functional harm, but it's unnecessary.
  • make consonant a local variable rather than a function parameter, as any inputted value is immediately clobbered with 0, and the result it returned by the function rather than written into consonant (which would be impossible as it is passed by value rather than passed by pointer/reference).

Upvotes: 8

Shadow2531
Shadow2531

Reputation: 12170

#include <iostream>
#include <string>
#include <cctype>
using namespace std;

size_t countConsonant(const string& s) {
    const string vowels("aeiou");
    size_t count = 0;
    for (string::const_iterator i = s.begin(); i != s.end(); ++i) {
        if (vowels.find(tolower(*i)) == string::npos) {
            ++count;
        }
    }
    return count;
}

int main() {
    cout << countConsonant("abcd") << endl;
}

Or, if you don't want non-alpha stuff to match, you can do it like this:

#include <iostream>
#include <string>
#include <cctype>
using namespace std;

size_t countConsonant(const string& s) {
    const string cons("bcdfghjklmnpqrstvwxyz");
    size_t count = 0;
    for (string::const_iterator i = s.begin(); i != s.end(); ++i) {
        if (cons.find(tolower(*i)) != string::npos) {
            ++count;
        }
    }
    return count;
}

int main() {
    cout << countConsonant("abcd") << endl;
}

Upvotes: 0

StackedCrooked
StackedCrooked

Reputation: 35505

How about this:

#include <algorithm>
#include <iostream>
#include <string>


bool isVowel(char c)
{
    switch (c)
    {
        case 'a':
        case 'e':
        case 'i':
        case 'o':
        case 'u':
        {
            return true;
        }
        default:
        {
            return false;
        }
    }
}


bool isConsonant(char c)
{
    return !isVowel(c);
}


int main(int argc, char *argv[])
{
    std::string test = "hello";
    std::size_t numVowels = std::count_if(test.begin(), test.end(), isConsonant);
    std::cout << "Number of vowels is: " << numVowels << std::endl;
    return 0;
}

Upvotes: 3

Xeo
Xeo

Reputation: 131829

First: && will test everything until something returns false or the end is reached. What you want is || which will test until a condition is true.
Second: What is shorter? Testing for consonants or testing for vowels? Vowels of course. Just put that test in an extra function (note: I assume a valid, alphabetical input string here):

function is_vowel(c) : bool
  for each vowel test
    if c == that vowel
      return true
  return false

After that, just replace your big conditional statement with a simple !is_vowel(str[i]). :) And last but not least, you want to increment your consonant variable, and there is a special operator for that: the increment operator! (Cool name, huh?)

++consonant; // better than consonant = consonant + 1; but same result

Upvotes: 2

Pete
Pete

Reputation: 10871

For starters you are using && (and) which is a logic error in this case. You want to use || (or).

You can also shorten the code a great deal by checking to see if they are NOT vowels.

int countConsonant(string str, int consonant)
{
    int length = str.length();
    int consonant = 0;

    for (int i = 0; i < length; i++)
    {
        if(!(str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' ||
            str[i] == 'u'))

        consonant = consonant + 1;
    }
    return consonant;
}

Someone else already posted an alternate form of this code as a function where you just pass in the variable you want to test.

Upvotes: 0

mrtsherman
mrtsherman

Reputation: 39882

Look into regular expressions. It allows you to specify a list of characters to be checked for. You would check each character in the string against the list and increment your counter.

http://www.johndcook.com/cpp_regex.html

Upvotes: 0

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