Beginner C: Reading Age - c

Question

I'm studying elementary programming in C and I'm doing a challenge to determine the reading age of various sentences. This is achieved by determining the amount of sentences in a string etc. I have some code that does the my first very basic step but it's not quite working as expected.
I'm thinking this is because my knowledge of the strlen function etc isn't sufficient.
I don't want to cheat for the answer as I like the sense of achievement from the problem solving. But would it be possible to get a gentle push in the right direction if at all possible?
char sentence[] = "One fish. Two fish. Red fish. Blue fish.";
int main(void)
{
int sentence_count = 0;
int word_count = 0;
int i;
int length = strlen(sentence);
for (i = 0; i == strlen(sentence); i++) //need to somehow go through a string one char at a time until the end.
{
if (sentence[i] == '.' || sentence[i] == '!' || sentence[i] == ';' || sentence[i] == '?')
{
return sentence_count ++;
}
if (sentence[i] == '\0')
{
return word_count ++;
}
}
printf("There are %i in %i sentences.\n", word_count, sentence_count);
printf("%i\n", length);
}

Answer 1

First Problem -
for (i = 0; i == strlen(sentence); i++)
This state should be -
for (i = 0; i < strlen(sentence); i++)
In your case, it would be terminating on the first iteration itself. However, You need to loop until you have reached the index - strlen(sentence).
Do recall that for loop has syntax - for(initialisation; condition; increment/decrement) will run only until the condition evaluates to true. So, you need the condition to evaluate to true till you have traversed the whole string which is done by the second line of code mentioned above.
A better alternative approach would be -
for (i = 0; sentence[i] != '\0'; i++)
which means the loop will run until you encounter a null-terminated character.
Second Problem -
return sentence_count ++;
.
.
return word_count ++;
Here, you don't need to add the return keyword before the above two statements. The return will directly exit from your program. Simply writing sentence_count++ and word_count++ would be correct.
Third Problem -
sentence[i] == '\0'
This statement doesn't quite fit with the logic that we are trying to achieve. The statement instead must check if the character is a space and then increment the word count -
if (sentence[i] == ' ')
{
return word_count ++;
}

Answer 2

Your for loop condition is the main problem; for (i = 0; i == strlen(sentence); i++) reads as "on entry, set i to 0, enter the body each time i is equal to the length of sentence, increment i at the end of each loop". But this means the loop never runs unless sentence is the empty string (has strlen of 0). You want to test i < strlen(sentence) (or to avoid potentially recomputing the length over and over, use the length you already calculated, i < length).
You also need to remove your returns; the function is supposed to count, and as written, it will return 0 the instant it finds any of the target characters, without using the incremented values in any way. Put a return 0; at the end of main to indicate exiting successfully (optionally, stdlib.h can be included so you can return EXIT_SUCCESS; to avoid magic numbers, but it's the same behavior).

Answer 3

The information in others answers being already covered, here are a couple of other suggestions for your consideration.
Remove function calls, such as strlen() from within the for(;;) loop. You've already obtained the length with:
int length = strlen(sentence);
Now, just use it in your for loop:
for(i = 0; i < length ; i++)//includes replacement of == with <
Encapsulate the working part of your code, in this case the the counting of words and sentences. The following uses a different approach, but its the same idea:
//includes deliminators for common end-of-sentence punctuation:
int count_sentences(const char *buf)
{
const char *delim = {".?!"};//add additional 'end-of-sentence' punctuation as needed.
char *tok = NULL;
int count = 0;
char *dup = strdup(buf);//preserve original input buffer
if(dup)
{
tok = strtok(dup, delim);
while(tok)
{
count++;
tok = strtok(NULL, delim);
}
free(dup);
}
return count;
}
One additional idea, that is out of scope with your original code, but very useful in practice is to remove any parts of the buffer that may not be part of the sentence, i.e. leading or trailing space. In your example, you have the test case for your sentences tightly defined within a string literal:
char sentence[] = "One fish. Two fish. Red fish. Blue fish.";
This is not incorrect, but what would happen at some point your code were to be expected to work with string buffers not so neatly packaged? i.e. leading or trailing white space characters in the buffer to be processed?
"\n\n\tOne fish. Two fish. Red fish. Blue fish.\f"
Removing unknown and unwanted content from the buffer prior to processioning simplifies the code doing the work, in this case counting sentences. Following is a simple example of how this can be done.
//prototype:
char *new = clear_leading_trailing_whitespace(sentence);
char * clear_end_space(const char *buf)
{
char *new = buf;
//clear leading whitespace
while (isspace(*new))
{
new++;
}
//clar trailing whitespace
int len = strlen(new);
while(isspace(*(new + len-1)))
{
len--;
}
*(new + len) = 0;
return buf;
}
Next, the following code segment is intending to count words:
if (sentence[i] == '\0')
{
return word_count ++;
}
But after being initialized to 0, word_count is only incremented once when seeing the null terminator, \0 once. Word count is generally a count of spaces between non-sentence terminating and non-whitespaces characters in a buffer. Or in otherwords, tracking how many clusters of non-whitespace there are. The following is a way to do this:
void countwords(const char *text, *count)
{
bool reading_word = false; // Flag
int words = 0;
for(int i=0; i<strlen(text); i++)
{
if(isspace(text[i])) {
reading_word = false;
}
else if(isalpha(text[i])) {
if(!reading_word) {
reading_word = true;
words++;
}
}
}
*count = words;
}
Functions like this can be used to greatly simplify contents of the main function:
char sentence[] = "One fish. Two fish. Red fish. Blue fish.";
int main(void)
{
int sentence_count = 0;
int word_count = 0;
char *new = clear_leading_trailing_whitespace(sentence);
countwords(new, &word_count);
sentence_count = count_sentences(new);
...
printf("There are %d words in %d sentences.\n", word_count, sentence_count);
}