additional logic to this exercise missing - objective-c

Question

Writing a basic program to count the number of words in a string. I've changed my original code to account for multiple spaces between words. By setting one variable to the current index and one variable to the previous index and comparing them, I can say "if this current index is a space, but the previous index contains something other than a space (basically saying a character), then increase the word count".
int main(int argc, const char * argv[]) {
#autoreleasepool {
//establishing the string that we'll be parsing through.
NSString * paragraph = #"This is a test paragraph and we will be testing out a string counter.";
//we're setting our counter that tracks the # of words to 0
int wordCount = 0;
/*by setting current to a blank space ABOVE the for loop, when the if statement first runs, it's comparing [paragraph characterAtIndex:i to a blank space. Once the loop runs through for the first time, the next value that current will have is characterAtIndex:0, while the if statement in the FOR loop will hold a value of characterAtIndex:1*/
char current = ' ';
for (int i=0; i< paragraph.length; i++) {
if ([paragraph characterAtIndex:i] == ' ' && (current != ' ')) {
wordCount++;
}
current = [paragraph characterAtIndex:i];
//after one iteration, current will be T and it will be comparing it to paragraph[1] which is h.
}
wordCount ++;
NSLog(#"%i", wordCount);
}
return 0;
}
I tried adding "or" statements to account for delimiters such as ";" "," and "." instead of just looking at a space. It didn't work...any idea what I can do, logically speaking, to account for anything that isn't a letter (but preferably just limiting it to these four delimiters - . , ; and space.

Answer 1

A standard way to solve these types of problems is to build a finite state machine, your code isn't quite one but its close.
Instead of thinking about comparing the previous and current characters think in terms of states - you can start with just two, in a word and not in a word.
Now for each state you consider what the current character implies in terms of actions and changes to the state. For example, if the state is not in a word and the current character is a letter then the action is increment word count and the next state is in a word.
In (Objective-)C you can build a simple finite state machine using an enum to give the states names and a case statement inside a loop. In pseudo-code this is something like:
typedef enum { NotInWord, InWord } State;
State currentState = NotInWord;
NSUInteger wordCount = 0;
for currentChar in sourceString
case currentState of
NotInWord:
if currentChar is word start character -- e.g. a letter
then
increment wordCount;
currentState = InWord;
InWord:
if currentChar is not a word character -- e.g. a letter
then
currentState = NotInWord;
end case
end for
The above is just a step from your original algorithm - recasting it in terms of states rather than the previous character.
Now if you want to get smarter you can add more states. For example how many words are there in "Karan's question"? Two. So you might want to allow a single apostrophe in a word. To handle that you can add a state AfterApostrophe whose logic is the same as the current InWord; and modify InWord logic to include if the current character is an apostrophe the next state is AfterApostrophe - that would allow one apostrophe in a word (or its end, which is also valid). Next you might want to consider hyphenated words, etc...
To test if a character is a particular type you have two easy choices:
If this is just an exercise and you are happy to stick with the ASCII range of characters there are functions such as isdigit(), isletter() etc.
If you want to handle full Unicode you can use the NSCharacterSet type with its pre-defined sets for letters, digits, etc.
See the documentation for both of the above choices.
HTH

Answer 2

I don't understand, You should be able to add or statements....
int main(void) {
char paragraph[] = "This is a test paragraph,EXTRAWORDHERE and we will be testing out a string.";
char current = ' ';
int i;
int wordCount = 0;
for (i = 0; i < sizeof(paragraph); i++){
if ((paragraph[i] == 32 || paragraph[i] == 44) && !(current == 32 || current == 44)){ //32 = ascii for space, 44 for comma
wordCount++;
}
current = paragraph[i];
}
wordCount++;
printf("%d\n",wordCount);
return 0;
}
I suppose it would be better to change the comparison of current from a not equal to into an equal to. Hopefully that helps.