Count number of consonants in a string - c

Question

Another starter question.
int counterConstant;
int x;
for(x = 0; x<20; x++){
if("bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSVWXYZ".IndexOf(tempString[x]) >= 0){
counterConsonant++;
}
}
But I get an error:
"error: member reference base type 'char [42]' is not a structure or union"
Is there another way I could do this?
(I'm doing this inside a for that checks each char on the string.)

Answer 1

There are no objects in C, so there are no "methods" and you can't call IndexOf on a string literal. A string is nothing more than an array of characters in C.
With that in mind, let's see how you can actually loop over the characters of a string:
for (const char *p = tempString; *p != '\0'; ++p) {
/* loop body */
char c = *p; // *p is the current letter
}
This will create a pointer to the first element of the string, and then loop over all of the following characters, if you'd really prefer to use indexes, you could do
for (size_t i = 0, len = strlen(tempString); i < len; ++i) {
char c = tempString[i];
}
As far as checking each letter for consonant-ness, that you can write a helper function for
int is_consonant(char c) {
c = tolower(c); // #include <ctype.h>
if (!isalpha(c)) return 0; // if not a letter, return false
switch (c) {
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
return 0;
default:
return 1;
}
}
now back to your loop, use this function to check each character.
int consonant_count = 0; // the =0 is important!
for (const char *p = tempString; *p != '\0'; ++p) {
if (is_consonant(*p)) {
++consonant_count;
}
}
If you don't initialize to 0, the initial value of consonant_count is unpredictable, so make sure you do.

Answer 2

If you are working on C (as it was specified in tags), strchr() method is used to search a char in a string, and strstr() is used to search a string in a string. We will use strchr() here because tempString[x] is a char. Also, don't forget to give your int variable an initial value. Try this code:
int main()
{
int counterConsonant = 0;
int x;
const char* tempString = "12345678901234567890";
for (x = 0; x<20; x++){
if (strchr("bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSVWXYZ", tempString[x]) != NULL){
counterConsonant++;
}
}
return 0;
}

Answer 3

C is a structured procedural language, so it doesn't have member functions/methods like a "true" object-oriented programming language such as C#. You could use a combination of strspn and strcspn like below to count sequences of consonants and non-consonant characters respectively, based on a predefined list of consonant characters:
#include <string.h>
size_t
count_consonants (const char *s)
{
size_t n;
size_t total = 0;
const char *consonants = "bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
/* While we haven't reached the end of the string,
execute the code in the body of the loop. */
while (*s != '\0')
{
/* Count the number of consonants starting at the current string position. */
n = strspn (s, consonants);
/* Add the number of consonants counted to
the total number of consonants found. */
total += n;
/* Advance the character pointer to the next character
that IS NOT a consonant, based on the number of consonants
stored in `n'. */
s += n;
/* Advance the character pointer to the next character
that IS a consonant (`strcspn' = skip the characters in
`s' that don't appear in `consonants'). */
s += strcspn (s, consonants);
}
return total;
}

Answer 4

char temp[20];
scanf("%s",temp);
int i,j, consonantsCounter=0;
char consonants[]={'b','c','d','f','g','h','j','k','l','m','n','p','q','r','s','t','v','w','x','y','z','B','C','D','F','G','H','J','K','L','M','N','P','Q','R','S','V','W','X','Y','Z'}
for(i=0;i<20;i++){
for(j=0;j<(sizeof consonants) / (sizeof consonants[0]);j++){
if(temp[i]==consonants[j]){
consonantsCounter++;
}
}
}