Question about String

[Marek]

I have question.. better yet problem with string... Since i just started with string.. i was wondering how to dissolve a word into letters... Like when i write for example FISH... it gives me result L1=F L2=I L3=S L4=H... and if u could kindly write also a explanation about it.

[hamsterman]

Ar you asking about c-strings or std::string ?
The general idea is the same for both though. A char can be accessed with []. example:

char str[] = "Hello world";
std::cout << "L" << 4+1 << "=" << str[4];//prints "L5=o" 

Then you have to iterate through the string (use for loop) and do this with every i, when i < string length.

[Null]

What do you have so far? This isn't hard - first you need to get the length of the string and then loop each character:

string str="aaa";
for(int i=0;i<str.length();i++)
  {
      cout <<str[i]<<endl; // print one character per line
     // ...
  }

Everything you need is here: http://www.cplusplus.com/reference/string/string/

EDIT: too slow

[Marek]

Thank you guys.. but i still need a bit of explanation....if it ain't a problem

for(int i=0;i<str.length();i++)

Correct me if im wrong... as i figured this
int i=0 is for Initialization
i<str.length() is the lenght of the word goes to zero
i++ every time it goes by one it gives a letter out

i know this is a bit newbie question but... i am confused about this part :D

[mcleano]

If in doubt, always consult the documentation: http://www.cplusplus.com/doc/tutorial/control/

The for loop, in this example, is saying:

1) Create and initialize a variable to 0.
2) For as long as i is less than the length of the string ( str.lenth() )
3) Increment i by one.

In Null's example, he is printing individual letters of the string "aaa" using str[i]. This means that on the first iteration the variable i will be 0 so he will be printing str[0]. On the second iteration, the variable i will be 1 so he will be printing str[1]. On the third etc...

[Null]

See http://www.cplusplus.com/doc/tutorial/control/#for
So yes, you are right.

[Marek]

Ok now i got it...thank you all ...I have one more question... with this code

for (i=0; i<(unos.length()); i++) {
	if ( (unos[i]=='a')||(unos[i]=='e')||(unos[i]=='i')||(unos[i]=='o')||(unos[i]=='u'|| 
		unos[i]=='A')||(unos[i]=='E')||(unos[i]=='I')||(unos[i]=='O')||(unos[i]=='U'))
		sa+=unos[i];

			}
	else
		su+=unos[i];

so basically i need to write out vowels but what bothers me is why doesn't it goes if( (i=='a')||(i=='e')...ect. but it goes if ( (unos[i]=='a')||(unos[i]=='e')...can somebody please explain it. (i found this on my schools web page).

[mcleano]

Because the variable 'i' is an integer, and not a char in the string object. The term unos[i] refers to the element i+1 in the string object. Making a comparison of if(i=='a') would be saying, for example, if(0=='a'); for the first iteration at least.

That code however is quite ugly. Try something more readable like:

   std::string str;
   const char vowels[] = { 'a', 'e', 'i', 'o', 'u' }; // this is 6 chars long

   for(int i=0; i<str.length(); ++i) {
      for(int x=0; x<6; ++x)
         if(!str[i]==(vowels[x]))
            // ...
         else
            // ...
   }