Function with array to find number of vowels and consonants

Hi all, I am trying to make a program capable to find with "*" the quantity of vowels and consonants that a input word may have.

i.e.

If I input the word first the output for finding vowels should be:

Code:

---

first
 *

---

And for consonants should be:

Code:

---

first
* ***

---

Hope you get an idea!
Thanks!

I have lots of ideas. But then, I don't actually care about finishing this program and getting a grade for it. So, what are your ideas?


Quzah.

As mentioned, tell us what you've done. In general, programming implementation details are usually straightforward. That is, once you know the problem and solution, its not very difficult to translate the solution to some programming language.

So, basically try and think how a person would solve this problem, for any given input. Because you have to consider any given input, you must give explicit details that will always work. This is basically what "pseudocode" is--an English description of the solution. So, whats your English description of the solution, or at least an attempt at it?

Find the standard library function that finds a character in a string. Establish a string of the vowels: "aeiou" (and sometimes 'y'?). Step through each letter in the input word and see if it's present in the vowel inventory. Print a '*' if yes.... or a space... or do the opposite print effect for showing consonants.

Quote:

---

Originally Posted by quzah

I have lots of ideas. But then, I don't actually care about finishing this program and getting a grade for it. So, what are your ideas?


Quzah.

---

Sorry I forgot to put what I have done by now:

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include "arrayv.h"

/* This function will find the number of vowels in the word and output their position using a * to
mark the vowels position. */
int arrayv()
{
        int i;
        char strv[11] = {'a','e','i','o','u','A','E','I','O','U','\0'};
        char str[100], repeat;
        printf("Number of vowels in a word:\n");
        do{
        printf("\n");
        printf("Please enter a word: ");
        scanf("%s", str);
        printf("\n");
        printf("%s\n", str);
        for(i=0; str[i] != '\0'; i++)
        {
                if (str[i] == strv[i])
                {
                        printf("*");
                }
                else
                {
                        printf(" ");
                }
        }
       
       
        printf("\n");
        printf("\n");
        printf("Play Again? ");
        getchar();
        repeat = getchar();
        while ( getchar() != '\n');
        } while(repeat == 'Y' || repeat == 'y');
       
        return EXIT_SUCCESS;
}

---

With this code, it works sometimes, but not always and is only vowels.

NOTE:It looks like the problem is where I marked, probably I'm not really writing the correct way the if statement.

Quote:

---

Originally Posted by nadroj

As mentioned, tell us what you've done. In general, programming implementation details are usually straightforward. That is, once you know the problem and solution, its not very difficult to translate the solution to some programming language.

So, basically try and think how a person would solve this problem, for any given input. Because you have to consider any given input, you must give explicit details that will always work. This is basically what "pseudocode" is--an English description of the solution. So, whats your English description of the solution, or at least an attempt at it?

---

Yeah, sorry for that. I forgot one of the main rules of this Forum.

Quote:

---

Originally Posted by nonoob

Find the standard library function that finds a character in a string. Establish a string of the vowels: "aeiou" (and sometimes 'y'?). Step through each letter in the input word and see if it's present in the vowel inventory. Print a '*' if yes.... or a space... or do the opposite print effect for showing consonants.

---

Thanks, it's sort like what I did, in exception that I can't find the standard library function to find the character in the string. As my guess is:

strchr()

Quote:

---

Originally Posted by georgio777

Code:

---

    for(i=0; str[i] != '\0'; i++)
    {
        if (str[i] == strv[i])
        {
            printf("*");
        }
        else
        {
            printf(" ");
        }
    }

---

---

Your checking index i of the string with index i of the "vowels". For example, the string "iae" you would check "i == a", "a == e", and "e == i", so it would report no vowels when in fact they are all vowels. Also, say the string entered was 15 chars, you will be comparing something like (among the other previous comparisons) "str[15] == strv[15]", which is obviously going past the bounds of the array "strv".
Basically, you need to check every character of your string, with every character of the "vowels". So you need two for loops, your outer one plus an inner one.

Quote:

---

Originally Posted by nadroj

Your checking index i of the string with index i of the "vowels". For example, the string "iae" you would check "i == a", "a == e", and "e == i", so it would report no vowels when in fact they are all vowels. Also, say the string entered was 15 chars, you will be comparing something like (among the other previous comparisons) "str[15] == strv[15]", which is obviously going past the bounds of the array "strv".
Basically, you need to check every character of your string, with every character of the "vowels". So you need two for loops, your outer one plus an inner one.

---

Oh, you are right, I'm comparing the same index of the input word with the one of the string index of the vowels.

My question is, in the for loops, both of them must include the if statement or the if statement is useless?

I was thinking as something similar to finding the biggest or smallest number of an array, in this case all the characters must be in the memory so they can be easily compared. Am I right?

Quote:

---

My question is, in the for loops, both of them must include the if statement or the if statement is useless?statement

---

You definitely need to compare "something", so the if statement is required and not useless. If you dont know if you need just one if statement inside the inner-most for loop, or if you need two (one inside the outer and inner for loop), then ask yourself what would either of these if statements be doing:
- if you put one in the outer for loop, what would it be checking?
- if you put one in the inner for loop, what would it be checking?
- if you put two (one in outer, one in inner) for loop, what would it be checking?

Hopefully you see that two of these three options dont really make sense, in which case the remaining choice is correct.

Quote:

---

I was thinking as something similar to finding the biggest or smallest number of an array, in this case all the characters must be in the memory so they can be easily compared. Am I right?

---

I dont know what you mean here, but theres no real tricks or magic you can do to solve this, so dont look for things like that. Use the method I suggested above/earlier.

Quote:

---

Originally Posted by nadroj

I dont know what you mean here, but theres no real tricks or magic you can do to solve this, so dont look for things like that. Use the method I suggested above/earlier.

---

I actually wouldn't use strchr. I mean you could, but I don't really see the point. Instead you'd be better served writing something like isvowel, and running through the string in a loop, and tagging either vowels or non-vowels. I'd probably just do something like:

Code:

---

for each character in the source string
    if this string's current character is  a vowel (or non vowel, depending what you want)
        set destination string (here) to be a star
    else
        set destination string (here) to be not-star
print source string and newline
print destination string and newline

---

Quzah.

Quote:

---

Originally Posted by nadroj

You definitely need to compare "something", so the if statement is required and not useless. If you dont know if you need just one if statement inside the inner-most for loop, or if you need two (one inside the outer and inner for loop), then ask yourself what would either of these if statements be doing:
- if you put one in the outer for loop, what would it be checking?
- if you put one in the inner for loop, what would it be checking?
- if you put two (one in outer, one in inner) for loop, what would it be checking?

Hopefully you see that two of these three options dont really make sense, in which case the remaining choice is correct.

I dont know what you mean here, but theres no real tricks or magic you can do to solve this, so dont look for things like that. Use the method I suggested above/earlier.

---

I fell really dumb since I don't really understand what are you trying to say.

I know that I need to do is nested for loops but I don't understand why. I sort got an idea that each character of the input word must pass thought the for loop so it can check if the character is or not a vowel, then when finished, continue with the second character and repeat the process until the word is finished.

But I don't know how to do that! :confused:

Quote:

---

Originally Posted by quzah

I actually wouldn't use strchr. I mean you could, but I don't really see the point. Instead you'd be better served writing something like isvowel, and running through the string in a loop, and tagging either vowels or non-vowels. I'd probably just do something like:

Code:

---

for each character in the source string
    if this string's current character is  a vowel (or non vowel, depending what you want)
        set destination string (here) to be a star
    else
        set destination string (here) to be not-star
print source string and newline
print destination string and newline

---

Quzah.

---

Oh, you're right, the teacher said something about isvowel, let see if I can do a little bit of more research related to that function!

Thanks!

Quote:

---

Originally Posted by georgio777

Oh, you're right, the teacher said something about isvowel, let see if I can do a little bit of more research related to that function!

Thanks!

---

They probably said you need to write one. There is no standard C 'isvowel' function.


Quzah.

Quote:

---

Originally Posted by quzah

They probably said you need to write one. There is no standard C 'isvowel' function.


Quzah.

---

Oh, ok. Probably I'm gonna stick with nadroj idea.

Quote:

---

Originally Posted by quzah

Quote:

---

Originally Posted by nadroj

I dont know what you mean here, but theres no real tricks or magic you can do to solve this, so dont look for things like that. Use the method I suggested above/earlier.

---

I actually wouldn't use strchr.

---

I never suggested to use any function, let alone "strchr". Your pseudocode is what I was suggesting earlier. I think you suggested to create a function which checks if a char is a vowel or not, and call it each time in the for loop. In that function, you would need to use a for loop. So both methods are essentially two nested for loops.

Quote:

---

Originally Posted by nadroj

I never suggested to use any function, let alone "strchr". Your pseudocode is what I was suggesting earlier. I think you suggested to create a function which checks if a char is a vowel or not, and call it each time in the for loop. In that function, you would need to use a for loop. So both methods are essentially two nested for loops.

---

Yes, and you are right, I'm not thinking in using any function or creating one.

I'm still trying to manage solving/writing the program with the two nested for loops.

Quote:

---

Originally Posted by nadroj

I never suggested to use any function, let alone "strchr".

---

Sorry, I confused you with the guy right below you.


Quzah.

No worries.

georgio777, my suggestion earlier and the pseudocode suggested by quzah do the same thing, one is directly using a nested for loop, the other is using a for loop and calling a function inside (which has a for loop)--either way they are nested for loops. So you can refer to his pseudocode earlier and just replace the function part with another for loop

Code:

---

for each character in string
  for each vowel
      if the string's character matches a vowel, its (obviously) a vowel and you can stop/break this for loop
  end for
  otherwise it is not a vowel
end for

---

Quote:

---

Originally Posted by nadroj

No worries.

georgio777, my suggestion earlier and the pseudocode suggested by quzah do the same thing, one is directly using a nested for loop, the other is using a for loop and calling a function inside (which has a for loop)--either way they are nested for loops. So you can refer to his pseudocode earlier and just replace the function part with another for loop

Code:

---

for each character in string
  for each vowel
      if the string's character matches a vowel, its (obviously) a vowel and you can stop/break this for loop
  end for
  otherwise it is not a vowel
end for

---

---

As my understand the code should look like this:

Code:

---

int i, j;

for(i=0; str[i] != '\0'; i++)
{
        for(j=0; strv[j] != '\0'; j++)
        {
                if (str[i] == strv[j])
                {
                printf("*");
                }
                else
                {
                printf(" ");
                }
        }
}

---

And it works only that there are HUGE spaces between each *

i.e.

OUTPUT

Code:

---

abeci
*                    *                    *

---

Its because your code doesnt match my pseudocode. I dont like to just give answers/solutions (which is why I waited to post any pseudocode), and I feel that the pseudocode already gave it away. If you look closely to every single thing in the pseudocode I posted, you should be able to spot whats different (another spoiler: do you know what the "break" statement does?)

Quote:

---

Originally Posted by nadroj

Its because your code doesnt match my pseudocode. I dont like to just give answers/solutions (which is why I waited to post any pseudocode), and I feel that the pseudocode already gave it away. If you look closely to every single thing in the pseudocode I posted, you should be able to spot whats different (another spoiler: do you know what the "break" statement does?)

---

Ok, you're right, I agree with you, it's better not to give the answers, so let's try again.
And yes I know what break does.

Code:

---

int i, j;

for(i=0; str[i] != '\0'; i++)
{
        for(j=0; strv[j] != '\0'; j++)
        {
                if (str[i] == strv[j])
                {
                printf("*");
                }
        break;
        }
        printf(" ");
}

---

In this case it only works on the first vowel.

You are printing a space every time, even if you find a vowel.


Quzah.

Quote:

---

Originally Posted by quzah

You are printing a space every time, even if you find a vowel.


Quzah.

---

Not with the break statement.
Even though I need to fix that.

Quote:

---

Originally Posted by georgio777

Not with the break statement.

---

Maybe if you actually put it in the right spot... but not the way you're doing it now. You actually need to see if you assign a vowel, and if so, don't put a space.


Quzah.

Quote:

---

Originally Posted by quzah

Maybe if you actually put it in the right spot... but not the way you're doing it now. You actually need to see if you assign a vowel, and if so, don't put a space.


Quzah.

---

That's why I'm trying to do, but I'm unsure about the break statement, because as I can see when the compiler passes through the breaking point, it stops the complete for loop.

Code:

---

for ...
    for ...
        if this then break
    this is the outer for, your break doesn't break from this loop

---

Therefore, you need to add another check in there, and if you've assigned a vowel, you need to not put a space.


Quzah.

Quote:

---

Originally Posted by quzah

Code:

---

for ...
    for ...
        if this then break
    this is the outer for, your break doesn't break from this loop

---

Therefore, you need to add another check in there, and if you've assigned a vowel, you need to not put a space.


Quzah.

---

When you said another check, you mean a if statement?

And the space is when is NOT a vowel.

For instance as I gave a example before:

Code:

---

first
 *

---

The * only appears on the vowels.


It's crazy I can't make it work :mad:!!!

You need to write it out, and then think about what you wrote, what each step means:

Code:

---

for each letter in the source string ... ok, how do I do that?
    if this letter is a vowel ... ok how do I do that?
        do a star or whatever ... ok how do I do that?
    else
        do the opposite of whatever (star/space) ... ok how do I do that?

---

You can do the first, you just use a for loop. Now, how do you do the second?


Quzah.

Quote:

---

Originally Posted by quzah

You need to write it out, and then think about what you wrote, what each step means:

Code:

---

for each letter in the source string ... ok, how do I do that?
    if this letter is a vowel ... ok how do I do that?
        do a star or whatever ... ok how do I do that?
    else
        do the opposite of whatever (star/space) ... ok how do I do that?

---

You can do the first, you just use a for loop. Now, how do you do the second?


Quzah.

---

Ok, let see:

for each letter in the source string ... ok, how do I do that?
#1 The source string is char str[100];
#2 Since is for each letter I need to create a for loop
#3 The int i; is the letter and the i++ is for each letter
#4 The str[i] != '\0' is for stopping the loop when the word is at the end of the string

The result:

Code:

---

for(i=0; str[i] != '\0'; i++)

---

if this letter is a vowel ... ok how do I do that?

#1 To find if the letter is a vowel I need to compare each character with the other vowel string
#2 To do that I need to create a nested for loop which is:

Code:

---

for(j=0; strv[j] != '\0'; j++)

---

#3 Where the int j; is the vowel, j++ is each vowel and the str[j] != '\0' is for stopping the loop when the word is at the end of the string
#4 Now each time the loop is working it should make the comparison with the vowels. For doing that you need to create an if statement in this case:

Code:

---

if (str[i] == strv[j])
{
        printf("*");
}

---

else do the opposite of whatever (star/space) ... ok how do I do that?

#1 For that I create an else statement:

Code:

---

else
{
        printf(" ");
}

---

So what's wrong?

The code should look like this:

Code:

---

        for(i=0; str[i] != '\0'; i++)
        {
                for(j=0; strv[j] != '\0'; j++)
                {
                        if (str[i] == strv[j])
                        {
                        printf("*");
                        }
                        else
                        {
                                printf(" ");
                        }
                }
        }

---

But doesn't work!

What happens if the letter isn't 'a', it's 'e'? Your if( this is a ) check fails, so you else print a space ... then you keep checking, and you see it's an e, and you print a *, then you keep checking, and you see it's not an i, so you print a space ...

This is why I suggested wrapping the inner for loop in a function that tells you if you've placed a vowel or not. Otherwise, you're going to still have to keep track of if you've successfully placed a * or not some other way.


Quzah.

Quote:

---

Originally Posted by quzah

What happens if the letter isn't 'a', it's 'e'? Your if( this is a ) check fails, so you else print a space ... then you keep checking, and you see it's an e, and you print a *, then you keep checking, and you see it's not an i, so you print a space ...

This is why I suggested wrapping the inner for loop in a function that tells you if you've placed a vowel or not. Otherwise, you're going to still have to keep track of if you've successfully placed a * or not some other way.


Quzah.

---

Oh, now I understand!

But what about this?

Why this one is not working also?

Code:

---

for(i=0; str[i] != '\0'; i++)
        {
                for(j=0; strv[j] != '\0'; j++)
                {
                        if (str[i] == strv[j])
                        {
                        printf("*");
                        }
                        break;
                }
                printf(" ");
        }

---

Quote:

---

Originally Posted by georgio777

The code should look like this:

Code:

---

    for(i=0; str[i] != '\0'; i++)
    {
        for(j=0; strv[j] != '\0'; j++)
        {
            if (str[i] == strv[j])
            {
            printf("*");
            }
            else
            {
                printf(" ");
            }
        }
    }

---

But doesn't work!

---

You only want to print a space if you know its a vowel. So the inner for loop is basically checking if its a vowel and if so, then printing a "*". In addition to printing some character, it should set a flag so that when youre outside of the inner for loop, you check this flag--if its true, it means a vowel was found so you dont print a space, otherwise a vowel was not found so you have to print a space. Since this flag must be visible within the inner for loop and within the outer for loop, it makes sense to declare this flag in the (start of the) outer for loop.

The crucial thing is some sort of "flag", saying "yes I did something (found vowel), so you dont have to do the other case (not vowel). As mentioned, you can achieve this by using a function (which has a for loop) and its return value is this implicit flag, or you directly use two for loops with an explicit flag.

Quote:

---

Originally Posted by nadroj

You only want to print a space if you know its a vowel. So the inner for loop is basically checking if its a vowel and if so, then printing a "*". In addition to printing some character, it should set a flag so that when youre outside of the inner for loop, you check this flag--if its true, it means a vowel was found so you dont print a space, otherwise a vowel was not found so you have to print a space. Since this flag must be visible within the inner for loop and within the outer for loop, it makes sense to declare this flag in the (start of the) outer for loop.

---

Yep, I understand reason why is not working and why is wrong, but how about the one you told me?

This one:

Code:

---

for(i=0; str[i] != '\0'; i++)
        {
                for(j=0; strv[j] != '\0'; j++)
                {
                        if (str[i] == strv[j])
                        {
                        printf("*");
                        }
                        break;
                }
                printf(" ");
        }

---

Maybe I forgot to mention the flag thing earlier, but that goes back to the fact that giving exact/complete answers/solutions isnt a good thing, you can try and figure out the last missing detail.

Quote:

---

Originally Posted by nadroj

The crucial thing is some sort of "flag", saying "yes I did something (found vowel), so you dont have to do the other case (not vowel). As mentioned, you can achieve this by using a function (which has a for loop) and its return value is this implicit flag, or you directly use two for loops with an explicit flag.

---

Well, that's what I am trying to do, but now I'm getting more and more lost :confused:

Quote:

---

Originally Posted by nadroj

Maybe I forgot to mention the flag thing earlier, but that goes back to the fact that giving exact/complete answers/solutions isnt a good thing, you can try and figure out the last missing detail.

---

At least give a hint. I don't really want to spent all day long trying to find the solution.

Another thing, I never heard of the definition flag or explicit flags, what does this means.

Code:

---

for...
    /* check for vowel */
    for ...
        if found a matching vowel
            yes we found a vowel
    /* outside of vowel check */
    did we find a vowel in there? if so, what does that mean for what we need to do?

---

You just need to work down through each step as you actually think them. Don't assume that thoughts like 'found a vowel' don't need their own step and that they're magically going to resolve themselves in your code. You need to actually think and translate each logical step, no mater how small it is, into some related action.


Quzah.

Quote:

---

Originally Posted by georgio777

Well, that's what I am trying to do, but now I'm getting more and more lost

---

Simplify: write a program that reads in a character and prints "vowel" if the character is a vowel, and "not vowel" otherwise.

Quote:

---

At least give a hint.

---

A few of us have given a number of hints already.

You can figure it out from my previous pseudocode

Code:

---

for each character in string
  for each vowel
      if the string's character matches a vowel, its (obviously) a vowel and you can stop/break this for loop
  end for
  otherwise it is not a vowel
end for

---

You can see that in the inner for loop theres an "if", and outside the inner for loop theres an "else". The only way this can be achieved is with some flag, i.e.

Code:

---

for each character in string
  flag = false;
  for each vowel
      if the string's character matches a vowel, its (obviously) a vowel and you can stop/break this for loop, set flag true
  end for
  if flag is false it is not a vowel so print space
end for

---

OK, I'm done with this, probably I'm going to talk directly with my teacher.

This is so confusing :confused:

Thank you anyway to everyone.

At least you gave me an idea!

Have a nice day and happy programming!

You giving up is just as frustrating for you as it is for us after devoting our time (at least I feel so).

The second piece of pseudocode in my previous post is pretty much the (one of the many) solution(s). Its a very slight modification of what you already have, just moving a thing or two, and adding a new variable and if statement.

Quote:

---

Originally Posted by nadroj

You giving up is just as frustrating for you as it is for us after devoting our time (at least I feel so).

The second piece of pseudocode in my previous post is pretty much the (one of the many) solution(s). Its a very slight modification of what you already have, just moving a thing or two, and adding a new variable and if statement.

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OK, I managed to make it work!!! Thank you for your time, I didn't know it was that easy. Sorry if I insisted to much. I accept I was a little immature. I promise it would happen again!

Again thank you to everyone, especially nadroj!

Here is the working code:

Code:

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#include <stdio.h>
#include <stdlib.h>
#include "arrayv.h"

/* This function will find the number of vowels in the word and output their position using a * to
mark the vowels position. */
int arrayv()
{
        int i, j, flag;
        char strv[11] = {'a','e','i','o','u','A','E','I','O','U','\0'};
        char str[100], repeat;
        printf("Number of vowels in a word:\n");
        do{
        printf("\n");
        printf("Please enter a word: ");
        scanf("%s", str);
        printf("\n");
        printf("%s\n", str);
        for(i=0; str[i] != '\0'; i++)
        {        flag = 0;
                for(j=0; strv[j] != '\0'; j++)
                {
                        if (str[i] == strv[j])
                        {
                        printf("*");
                        flag = 1;
                        break;
                        }
                }
                if (flag == 0)
                {
                printf(" ");
                }
        }
        printf("\n");
        printf("\n");
        printf("Play Again? ");
        getchar();
        repeat = getchar();
        while ( getchar() != '\n');
        } while(repeat == 'Y' || repeat == 'y');
       
        return EXIT_SUCCESS;
}

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Quote:

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Originally Posted by georgio777

I promise it would happen again!

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Hey now!

Quote:

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Originally Posted by nadroj

Hey now!

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Hahahahahahaha

Sorry

I promise it would NOT happen again!