please give me the algorithm or pseudocode

[nesir]

i want to make a program that will ask a sentence from the user and it will display the number of vowels found in the sentence.

example...
Enter sentence: i am a newbie of c programming.

there are 3 a. there are 3 i. there are 0 u.
there are 2 e. there are 1 o.

help anyone? just the correct commands to type and i'll supply the others... PLEASE!!!!! ONEGAI!!!

[Dave_Sinkula]

Loop through each character of the string, test each character with against each vowel -- 'a', 'e', 'i', 'o', or 'u', if the character is the vowel, increment an associated counter that had been initialized originally to zero.

Do you know how to input a string? You may want to check the FAQ.

[nesir]

please give me somthing more specific... like waht are the commands i'm going to use... and if possible, show me the code.. :-)

[valis]

You will be using these keywords and operators: int main return char * [] for ++
You will need these functions: puts fgets printf

I think that's everything.
Anyone see something I missed?

[VirtualAce]

Better idea. You show us some code and we will tell you where you screwed up.

[Moony]

Hope this help

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
   int c,u,a,e,o,i;
   u=a=e=o=i=0;
   char text[100];
   printf("Enter text please\n ");
   gets(text);
   for (c=0; c<=strlen(text); c++)
   {
     if ( text[c] =='u' || text[c] =='U') u++;
     else if (text[c] == 'a' || text[c] =='A') a++;
     else if (text[c] == 'e' || text[c] =='E') e++;
     else if (text[c] == 'o' || text[c] =='O') o++;
     else if (text[c] == 'i' || text[c] =='I') i++;

   }
   printf ("The result:\n");
   printf ("There are %d u\n",u);
   printf ("There are %d a\n",a);
   printf ("There are %d e\n",e);
   printf ("There are %d o\n",o);
   printf ("There are %d i\n",i);
   return 0;
}

Example:

Enter text please
I am a newbie of c programming
The result:
There are 0 u
There are 3 a
There are 2 e
There are 2 o
There are 3 i

[Salem]

Using gets() - read the FAQ !!!
http://faq.cprogramming.com/cgi-bin/...&id=1043284351

[itsme86]

Algorithmically speaking, this is about the worst way you could control a loop iterating through a string:

for (c=0; c<=strlen(text); c++)

strlen() has to count the number of characters in the string every time you iterate through the loop. Using text[c] to test if you've reached the nul-terminator is a much better control method.

[Moony]

Using gets() - read the FAQ !!!
http://faq.cprogramming.com/cgi-bin/...&id=1043284351

If it is bad buffer why it is still there and in use?

[Moony]

Algorithmically speaking, this is about the worst way you could control a loop iterating through a string:

for (c=0; c<=strlen(text); c++)

strlen() has to count the number of characters in the string every time you iterate through the loop. Using text[c] to test if you've reached the nul-terminator is a much better control method.

Good to know, thanks.

[Salem]

> If it is bad buffer why it is still there and in use?
Huh?
Do not use gets(), EVER
It's a fairly simple rule to obey.

[Richie T]

>>why it is still there

http://en.wikipedia.org/wiki/Gets

>>and in use?

Well because people like yourself learn it and use it of course! I highly doubt that anyone
getting paid to write code is using it, and any code it does use it is probably very, very old.

gets: kind of like leaving the house, but locking yourself out - you cant get back in!

[Moony]

Thanks all. What about scanf()?

[itsme86]

What about scanf() indeed. Are you asking if it's unsafe? If you misuse it it can be. Being smart/careful with the format string will keep it safe, though.

[fgw_three]

Thanks all. What about scanf()?

Do a google on "secure programming in C" and you'll probably find a bunch of useful information on which C functions can be misused by the programmer and subsequently abused by the bad guys.