Psuedo-code for Count Words problem

CountWords.psuedo

// Test Cases:
// “A BIG DOG” should return 3
// “A” should return 1
// “  BC “ should return 1

int CountNumWordsInSentence(const char* InString) {
	bool wasLetter = true;
	int wordCount = 0;
	
  // For this for loop. take advantage that InString is a char pointer.
  // This means that each character in the string points to the next one, 
  // making easy to iterate through the letters.
  // Lookup how to iterate through letters of a char pointer string
	for(// each char in InString) {
		charIsLetter = isLetter(char);
		if (!wasLetter & isLetter) {
			wordCount++;
		}
		wasLetter = isLetter;	
	}
}

bool isLetter(char* letter) {
  // a char in c is actually just an ASCII character code
  // these codes go in order and go from 65 to 90 for uppercase and 97 to 122
  // so you can see if letter is between those values or just do
  // (letter >= 'a' && letter <= 'z') || (letter >= 'A' && letter <= 'Z')
	if ( // letter is A-Z ) {
		return true;
	}
	return false;
}