How to read spaces, words, and letters line-by-line from a file in a function using Loops (C )-CodePudding

so I am having a problem with getting my program to increment values properly.

My program needs to take a file in main(), and then pass that to a function-set to print that is called in main.

The key thing is that I need to use loops within the functions to get Letter Count, Space-Count, and Word Count.

I have the output configured right

cout << line_number << ": " << line << " [" << letter_count << " letters, " << space_count << " spaces, " << word_count << " words]" << endl;

Which results for example

0: Words go here. [# letters, # spaces, # words.]

But with my current functions for Letters and spaces, it doesn't work.

My non-space function for example

int count_non_space(string line) {
    int non_space = 0;
    for (int i = 0; i < line.length(); i  ) {
        if (line.c_str() != " ") {
            non_space  ;
        }
    }
    return non_space;

It counts all of the characters in the line instead and the counterpart (space_count) counts nothing.

And that's not to mention that I don't know how to count the words in the line.

Any advice as to what is going on? as I am certain that count_space and count_non_space should be inverses of each other (count_space being the same function but with == instead of !=)

EDIT: Got the Letter and Space count correct. Now, how would I get the word count from that sort of method?

EDIT 2: Okay so letter count is off. It is counting puncutation-characters (commas, periods, dashes, hiphons.etc) as leters.

I have managed to redact periods, dashes.etc from the code manually with a reduction if statement in the count_non_characters function.

But I can't add ' to it as it already uses '' to catch the char comparison

Is there catch-all term for punctuation characters in C that I can use for

if (line[i] == "Puncutation") {
   non_space--;
}

CodePudding user response：

As UnholySheep said, when you compare a c string (char *) you can't use standard logical operators. You will need to use strcmp(). However, if you use a c std::string then you can use compare() or logical operators.

As for finding words in a string. Here are a few resources.

For further help. Google: "Get word count per line c "

Reminder, these two are different data types and have different library support:

std::string myStr
myStr.c_str()

CodePudding user response：

If the goal is to count characters in a string that are not spaces, then there is a way to do this using the STL and lambdas that is much cleaner than writing a bunch of loops and worrying about updating variables.

int count_non_space(std::string line) {
    return std::count_if(line.begin(), line.end(),
                         [](auto ch) { 
                             return ch != ' '; 
                         });
}

This also makes is very straightforward to accommodate for things like spaces and tabs.

int count_non_space(std::string line) {
    return std::count_if(line.begin(), line.end(),
                         [](auto ch) { 
                             return ch != ' ' && ch != '\t'; 
                         });
}

To count the opposite (just the spaces) we simply need to change the condition in the lambda.

int space_count(std::string line) {
    return std::count_if(line.begin(), line.end(),
                         [](auto ch) { 
                             return ch == ' ' || ch == '\t'; 
                         });
}

As Remy Lebeau helpfully points out, we don't even have to write the lambda. We can simply use the std::isspace function directly instead of the lambda.

int space_count(std::string line) {
    return std::count_if(line.begin(), line.end(), std::isspace);
}

Documentation on std::count_if.

CodePudding user response：

Here's how I would revise the function you gave:

int count_non_space(string line) {
    int non_space = 0;
    for (int i = 0; i < line.length(); i  ) {
        if (line[i] != ' ') {
            non_space  ;
        }
    }
    return non_space;
}

Notes:

I changed line.c_str() to line[i] in order to access the ith character of line
I changed " " to ' ' so that it's comparing against the space char, not a string which only contains the space. The comparison would fail if we were comparing the ith char to a string

As for this:

And that's not to mention that I don't know how to count the words in the line.

I don't know how your requirements define a word, but if we assume a word is any contiguous clump of non-space characters, you could use this logic:

initialize bool in_word to false
initialize int word_count to 0
for each char in the string:
1. if in_word is false and the current char is not a space, then set in_word to be true and increase word_count by 1
2. if in_word is true and the current char is a space, then set in_word to be false
return word_count