Can you lend me a hand with this word counting code? - c

Question

This code don't count words properly. I don't know if it is wrong on the for or what. Need help.
#include <stdio.h>
#include <stdlib.h>
int count_p(char sentence[100]) {
int i, m = 1;
for (i = 0 ; i < 100 ; i++) {
if (sentence[i] == ' ') {
m += 1;
}
}
return(m);
}
void main() {
char s[100];
int p;
printf("Sentence here: ");
scanf("%s", &s[50]);
p = count_p(sentence);
printf("Words: %d", p);
printf("\n");
}

Answer 1

The %s in scanf stops reading when it found a whitespace. Therefore, ' ' won't appear in s unless it was there as indeterminate value in uninitialized variable.
You can use fgets to read a whole line.
Here is a fixed code that also checks for end of the string.
#include <stdio.h>
#include <stdlib.h>
int count_p(char sentence[100]) {
int i, m = 1;
for (i = 0 ; i < 100 && sentence[i] != '\0'; i++) {
if (sentence[i] == ' ') {
m += 1;
}
}
return(m);
}
int main(void) {
char s[100];
int p;
printf("Sentence here: ");
fgets(s, sizeof(s), stdin);
p = count_p(s);
printf("Words: %d", p);
printf("\n");
return 0;
}

Answer 2

scanf("%s", &s[50]);
Not a correct way to take input and writing at index which is out of bound. Do this instead -
scanf("%99[^\n]", s); // this will read 99 characters and until '\n' is encountered
In main you function call is incorrect -
p = count_p(sentence); // sentence is not declares in main
Call like this -
p = count_p(s); // pass s instead of sentence to function
Also in function count_p change ccondition in for loop as -
size_t i;
size_t len=strlen(s);
for (i = 0 ; i < len ; i++)

Answer 3

You see &s[50] means that you pass a pointer to the 51-th element of s, you then try to access s from the beginning but, the first 50 characters in s were not yet initialized, this leads to undefined behavior.
Also, your loop from 0 to 99 will have the same issue since you might input a string of less than 100 characters, in that case you would be accessing uninitialized data too.
You can fix your program by changing this
scanf("%s", &s[50]);
to
scanf("%99s", s);
and then
for (i = 0 ; i < 100 ; i++) {
to
for (i = 0 ; s[i] != '\0' ; i++) {
because scanf() will append a '\0' to make the array a valid c string, that's also the reason for the "%99s".
Another problem is that, if you want white space characters not to make scanf() stop reading, you need a different specifier, because "%s" stops at the first white space character, this is a suggestion
scanf("%99[^\n]", s);
Or you can do as #MikeCAT suggested and go with fgets(). But be careful with the trailing '\n' in case of fgets().
And finally, altough highly unlikely in this situation, scanf() might fail. To indicate success it returns the number of specifiers actually matched, thus it might indicate partial success too. It's fairly common to see the return value of scanf() ignored, and it's very bad when you have a "%d" specifier for example because then the correspoinding parameter might be accessed before initializing it.

Answer 4

The statement scanf("%s", &s[50]); is in correct in your situation.Since you want to enter a sentence separated by spaces,the correct way of doing it is :
scanf(" %99[^\n]s",sentence);
That will prevent buffer overflow and allow space between words.Also your program does not seem to count words correctly if the sentence has consecutive whitespaces.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int count_p(char *sentence);
void main()
{
char sentence[100];
printf("Sentence here: ");
scanf(" %99[^\n]s",sentence);
int p = count_p(sentence);
printf("Words: %d", p);
printf("\n");
}
int count_p(char *sentence)
{
int len = strlen(sentence);
int x = 0 , wordCount = 0;
for( int n = 0 ; n < len ; n++ )
{
x++;
if( sentence[n] == ' ' )
x = 0;
if( x == 1 )
wordCount++;
}
return wordCount;
}