C Code to count how many letters in a word - c

Question

I made simple code to count how many letters in a word, for some reason the variables letters is not consistent, here is the code
#include <stdio.h>
int main()
{
char str;
int slen(char *str)
{
int i;
int letters=0;
for(i=0;i<30;i++)
{
if((str[i]>='A'&&str[i]<='Z')||(str[i]>='a'&&str[i]<='z'))
letters++;
}
return letters;
}
printf("%d \n", slen("Word"));
}
The variable letter here prints 22.

Answer 1

Because of the < 30 condition in your for loop, your program goes past the "Word" string boundaries and starts calculating whatever's in memory right after that string.
Replace < 30 with < strlen(str) (you're going to need to #include <string.h>).
UPDATE: (kudos to #J...S (seee comments)) alternatively you can rewrite
for (i=0; i < 30; i++)
{
if((str[i]>='A'&&str[i]<='Z')||(str[i]>='a'&&str[i]<='z'))
letters++;
}
with
while (str != '\0') {
if ((*str >= 'A' && *str <= 'Z') || (*str >= 'a' && *str <= 'z')) {
letters++;
}
++str;
}
You could also use something like isalpha() instead of your if condition.
And, by the way, please use spaces to separate different parts of a line. Otherwise it's really hard to read. Thanks :)

Answer 2

The function should be declared/defined outside main().
Change this:
for(i=0;i<30;i++)
to this:
for(i = 0; i < strlen(str); i++)
because you want to iterate over the length of the string, and not more or less than its actual length.
In your case, "Word" had just 4 letters, and you iterated until i < 30 evaluated to false, invoking Undefined Behavior, since you would access memory that is out of bounds.
Putting everything together, we get:
#include <stdio.h>
#include <string.h>
int slen(char *str)
{
int letters=0;
for(unsigned int i=0;i<strlen(str);i++)
{
if((str[i]>='A'&&str[i]<='Z')||(str[i]>='a'&&str[i]<='z'))
letters++;
}
return letters;
}
int main()
{
printf("%d \n", slen("Word"));
}
Output:
4

Answer 3

You want this:
#include <stdio.h>
#include <string.h>
// declaring slen outside main (not using non standard GNU extension)
int slen(char *str)
{
int i;
int letters = 0;
int length = strlen(str); // using strlen, in order not to go beyond the end
// of the string
for (i = 0; i < length; i++) {
if ((str[i] >= 'A' && str[i] <= 'Z') || (str[i] >= 'a' && str[i] <= 'z'))
letters++;
}
return letters;
}
int main()
{
printf("%d \n", slen("Word123"));
}
Expected output here is 4 (4 lettres of "Word", not counting digits 1,2, and 3).
Slighly optimized version:
int slen(char *str)
{
int letters = 0;
for (int i = 0; ; i++) {
char c = str[i];
if (c == 0)
break;
if ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z'))
letters++;
}
return letters;
}
and even better using isalpha from ctype.h:
#include <ctype.h>
...
int slen(char *str)
{
int letters = 0;
for (int i = 0; ; i++) {
int c = str[i]; // yes, int is correct, isalpha wants an int
if (c == 0)
break;
if (isalpha(c))
letters++;
}
return letters;
}

Answer 4

For starters according to the C Standard the function main without parameters shall be declared like
int main( void )
Though some compilers can have their own language extensions nevertheless again according to the C Standard you may not define a function inside another function.
Also take into account that it is not necessary that Latin letters follow each other sequentially in a given character set. For example in the EBCDIC set some letters do not follow sequentially each other. There are embedded other symbols between letters.
So it is better to use standard C function isalpha declared in the header <ctype.h>.
Also the original string is not changed in the function. So it should be declared with the qualifier const.
And it is unclear why this magic number 30 is used in this loop within the function
for(i=0;i<30;i++)
Taking all this into account the program can look like
#include <stdio.h>
#include <ctype.h>
size_t letters_count( const char *s )
{
size_t count = 0;
for ( ; *s; ++s )
{
if ( isalpha( ( unsigned char )*s ) ) ++count;
}
return count;
}
#define N 100
int main(void)
{
char s[N];
printf( "Enter a statement: " );
fgets( s, sizeof( s ), stdin );
printf( "The statement contains %zu letters\n", letters_count( s ) );
return 0;
}
Its output might look like
Enter a statement: Hello TheErolley
The statement contains 15 letters