Counting Occurrences Of A Word In A String | C Programming Example

[jhunecamatura]

Thank you. This is the exact problem we faced during our lab meeting

[theranker]

Really like your explanations and tutorials. Just finished your C Programming course a couple of days ago and trying out these examples now. I have however, spotted a problem in this code.

You're using 'i' to iterate through the string and 'j' to iterate through the word. This approach is great as long as you do not have a word in your string which has the word that you are searching as a part.

For example,
if the string is "Notwithstanding a brilliant defence, he was found guilty and charged with a second degree offence."
If the word being searched for is "with" then the program will display the word count to be 2.

Anyways, hope you can find a solution to counter this problem in future videos :)

[jsnam]

omg sir you saved my life !! Subscribed immediately! Much thanks!

[juju]

does the function only proceed to the nested for loop when there is a character match?

[MrJzvoyeur]

one remark here: when a word is found we may do: j += (wlen-1);
can the word be found just one position after its 1st occurrence in the string?
let the string be and the word be "aa".
what result would we expect? 3 or 5?

[naboulsikhalid]

hi, good tutorials that make me grow. thanks a lot. I have an issue I used this sentence: "It is is is a different word this time", to count "is", it shows count = 4. is this correct?Can you help

[SketchupGuru]

I don't understand why you added the Asterix to the function?
Why did you make the string a pointer?
Isn't it possible to run the function without making it a pointer

[ayzikdig]

why the +1 int the
int end? is it for the '/o' ?

[wxlfxdc]

How to count the occurence of a word without typing the specific word?

[mohammedmad]

Here's the corrected version of the sentence:

Thank you for this amazing explanation, But we want to know how we can deal with uppercase and lowercase letters and some special characters in a string. I know how to convert lower and upper characters using an ASCII code, but when it comes to special characters like ('), or even spaces as you said before, I have no idea how to work with them.

[__iTsMe__]

before i watch the video i came with this solution:
int word_count(char *str, const char *word)
{
int word_len = strlen(word), count = 0, i = 0, j = 0;
while (str[i] != '\0')
{

if ( {tolower(str[i]) - tolower(word[j]) == 0)
{
j++;
if (j == word_len)
{
count++;
j = 0;
}
}
i++;
}
return count;
}
is an efficient solution or not...?
can i have feedback on this function..?

[nishka]

Can you please do this program using StringTokenizer?

[sanjayagihan]

When the word is "and" & string is "i understand"

The output shoud be zero
but it will print 1 because of "and" inside the word "understand" how to overcome this error

[xturki]

i thing in strtok function it will be easier but still Great explanation

[justcurious]

Amazing, Thanks :
// Full code
int word_count(char *string, char *word){
int slen = strlen(string);
int wlen = strlen(word);
int end = slen - wlen + 1;
int count = 0;
for(int i = 0; i < end; i++){
bool word_found = true;
for(int j = 0; j < wlen; j++){
if(word[j] != string[j+i]){
word_found = false;
break;
}
}
if(word_found){
++count;
}
}
return count;
}

[deepblackoutlaw]

the way i solved it before watching the video:
int word_count(const char* str, const char* word) {
size_t str_length = strlen(str);
size_t word_length = strlen(word);
int counter = 0;

for (size_t i = 0; i < str_length; i++) {
if (*(str + i) == *(word + 0)) {
size_t j = 1;
for (j; j < word_length; j++) {
if (*(str + (i + j)) != *(word + j)) {

}
}

i += j;
if (j == word_length) {
counter++;
}
}
}

return counter;
}

[aminetakha]

Hello sir, I come up with the following solution. Do you think it is efficient

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
char string[] = "world worlhdworld";
char word[] = "world";
char *sub = malloc(sizeof(word));
int count = 0;

for (int i = 0; i < strlen(string); i++)
{
if (string[i] == word[0])
{
strncpy(sub, string + i, strlen(word));
int is_equal = strcmp(sub, word);
if (is_equal == 1)
{
count++;
i += strlen(word) - 1;
}
}
}
free(sub);
return 0;
}

[MrJzvoyeur]

one remark (or question) here:
when a word is found we may do:
if (word_found) { count++; i += (wlen-1); }
which means: a word of more than one character should not
be found just one position after its 1st character in the string.
let the string be and the word be "aa" .
the result of counting should be: 3, not 5, right?

[Abdalrahman_]

--> This is my solution before watching your video

#include <stdio.h>
#include <string.h>

int is_found(char string[], char word[]){
int count=0, i;
char W[20];
for(i=0; string[i]; i+=(strlen(W)+1) ){
sscanf(string+i, "%[^' ']", W);
if( !strcmp(word, W) )
count++;
}
return count;
}

int main(void) {

char string[]= "Hi Hello Society Hi Hi Hi Hello Hello Society";
char word[10];
printf("What's the word that u search for: ");
scanf("%s", word);

printf(is_found(string, word)? "Found %d times\n" : "Not found\n", is_found(string, word));

return 0;
}