Why is this still counting spaces within a String? - c

Question

I'm just starting to code and I need help figuring out why this loop counts spaces within a string.
To my understanding, this code should tell the computer to not count a space "/0" and increase count if the loop goes through the string and it's any other character.
int main(void)
{
string t = get_string("Copy & Past Text\n");
int lettercount = 0;
for (int i = 0; t[i] != '\0'; i++)
{
lettercount++;
}
printf("%i", lettercount);
printf("/n");
}

Answer 1

\0 represents the null character, not a space. It is found at the end of strings to indicate their end. To only check for spaces, add a conditional statement inside the loop.
int main(void)
{
string t = get_string("Copy & Past Text\n");
int lettercount = 0;
for (int i = 0; t[i] != '\0'; i++)
{
if (t[i] != ' ')
lettercount++;
}
printf("%i", lettercount);
printf("\n");
}

Answer 2

Space is considered a character, your code goes through the string (an array of characters) and counts the characters until it reaches the string-terminating character which is '\0'.
Edit: set an if condition in the loop if(t[i] != ' ') and you wouldn't count the spaces anymore.

Answer 3

You misunderstand the nature of C strings.
A string is an array of characters with a low value ( '\0') marking the end of the string. Within the string some of the characters could be spaces (' ' or x20).
So the " t[i] != '\0' " condition marks the end of the loop.
A simple change:
if ( t[i] != ' ') {
lettercount++;
}
Will get your program working.

Answer 4

This for loop
for (int i = 0; t[i] != '\0'; i++)
iterates until the current character is the terminating zero character '\0' that is a null character. So the character is not counted.
In C there is the standard function isalpha declared in the header <ctype.h> that determines whether a character represents a letter.
Pay attention to that the user can for example enter punctuation symbols in a string. Or he can use the tab character '\t' instead of the space character ' '. For example his input can look like "~!##$%^&" where there is no any letter.
So it would be more correctly to write the loop the following way
size_t lettercount = 0;
for ( string s = t; *s; ++s )
{
if ( isalpha( ( unsigned char )*s ) ) ++lettercount;
}
printf("%zu\n", lettercount );
This statement
printf("/n");
shall be removed. I think instead you mean
printf("\n");
that is you want to output the new line character '\n'. But this character can be inserted in the previous call of printf as I showed above
printf("%zu\n", lettercount );

Answer 5

A null-terminator is the last leading element in a character array consisting of a string literal (e.g. Hello there!\0). It terminates a loop and prevent further continuation to read the next element.
And remember, a null-terminator isn't a space character. Both could be represented in the following way:
\0 - null terminator | ' ' - a space
If you want to count the letters except the space, try this:
#include <stdio.h>
#define MAX_LENGTH 100
int main(void) {
char string[MAX_LENGTH];
int letters = 0;
printf("Enter a string: ");
fgets(string, MAX_LENGTH, stdin);
// string[i] in the For loop is equivalent to string[i] != '\0'
// or, go until a null-terminator occurs
for (int i = 0; string[i]; i++)
// if the current iterated char is not a space, then count it
if (string[i] != ' ')
letters++;
// the fgets() reads a newline too (enter key)
letters -= 1;
printf("Total letters without space: %d\n", letters);
return 0;
}
You'll get something like:
Enter a string: Hello world, how are you today?
Total letters without space: 26
If a string literal has no any null-terminator, then it can't be stopped from getting read unless the maximum number of elements are manually given to be read till by the programmer.