Here is my function to remove the vowels in a string;
char *removeVowels(char *inString) {
int count = 0; //to count the non vowel characters
for (int i = 0; inString[i]; i++)
if (inString[i] != 'a' || inString[i] != 'e' || inString[i] != 'u' || inString[i] != 'o' || inString[i] != 'i')
inString[count++] = inString[i]; //if character is not a vowel placed at count++
inString[count] = '\0';
return inString;
}
The problem is that it is returning the original string inputted. Any ideas?
There is a confusion between the || and && operators. You want to test that the character is different from 'a' AND different from 'e' etc.
Here is a modified version:
char *removeVowels(char *inString) {
int count = 0; // index for preserved characters
for (int i = 0; inString[i]; i++) {
if (inString[i] != 'a' && inString[i] != 'e' && inString[i] != 'i'
&& inString[i] != 'o' && inString[i] != 'u') {
inString[count++] = inString[i]; // copy the non-vowel character
}
}
inString[count] = '\0'; // set the null terminator.
return inString;
}
Note however that uppercase vowels are not removed by this function, and whether y should be considered a vowel remains to be decided.
As stated in another comment, you need to use && instead of || to make sure that the character does not match any vowels. It might be easier to create a new string and add non-vowels to that string as you go. Something like:
char *removeVowels(char *inString, int size){
char newString[size];
int count = 0;
for(int i = 0; i < size; i++){
if(inString[i] != 'a' && inString[i] != 'e' && inString[i] != 'i' && inString[i] != 'o' && inString[i] != 'u'){
newString[count] = inString[i];
count++;
}
}
newString[count] = '\0';
return newString;
}
Related
I have to find the vowels in an input string and return the string of the vowels in reverse order.
Example input: "woUIW"
The expected output is "IUOo"
But I cannot use pointers, only recursion.
This is my attempt:
void letters_revers(char s[])
{
int i = 0,g,len,right,left;
char reversed[10];
len=strlen(reversed)-1; // get the length of the string
left = 0; // set left index at 0
right = len - 1; // set right index len - 1
while(i!='\0')
{
if (s[i] == 'a' || s[i] == 'A' || s[i] == 'e' || s[i] == 'E' || s[i] == 'i' || s[i] == 'I' || s[i] =='o' || s[i]=='O' || s[i] == 'u' || s[i] == 'U')
{ reversed[i]+=s[i];
for(g = len - 1; g >= 0; g--)
{
printf("%c", reversed[g]);
}
}
}i++;
}
I am getting an int of the entire string s for 'letter', the conditions in my 'if' statement seem to not be reading properly - is my syntax incorrect?
I get user input:
string s = get_string("Text here: ");
the function is as follows:
int letter_count(string s)
{
int i =0;
int len = strlen(s);
int letter = 0;
while(i < len)
{
if (s[i] != '\0' || s[i] != '.' || s[i] != ',' || s[i] != '!' || s[i] != '?')
{
letter++;
}
i++;
}
return letter;
}
then call the function:
int letter = letter_count(s);
printf("letter Count: %i\n", letter);
Try changing the OR operator with the AND
if (s[i] != '\0' || s[i] != '.' || s[i] != ',' || s[i] != '!' || s[i] != '?')
is ALWAYS true. Because any character is either not "." or not ",". Which letter would you expect to be both?
You want to check whether the current letter is "not ." AND "not ," AND "not !".
I.e.
if (s[i] != '\0' && s[i] != '.' && s[i] != ',' && s[i] != '!' && s[i] != '?')
Almost correct, you have to change the type of the argument to char*, there is no string type on the default libraries. (Documentation of string library).
Working example with the modifications:
#include <stdio.h>
#include <string.h>
int letter_count (char* s)
{
int i = 0;
int len = strlen (s);
int letter = 0;
while (i < len)
{
if (s[i] != '\0' && s[i] != '.' && s[i] != ',' && s[i] != '!'
&& s[i] != '?')
{
letter++;
}
i++;
}
return letter;
}
int main ()
{
char my_word[] = "Sample word";
printf ("'%s' have %d letters",my_word, letter_count (my_word));
return 0;
}
Output:
'Sample word' have 11 letters
I am trying to create a program that removes the vowels from a sentence. However, my program keeps failing because it keeps printing zero byte in the string. Would anyone be able to show me where I am going wrong?
#include <stdio.h>
int remove_all_vowels(int character);
int main(void) {
int character = getchar();
while (character != EOF && character != '\0') {
int new_character = remove_all_vowels(character);
putchar(new_character);
character = getchar();
}
return 0;
}
int remove_all_vowels(int character) {
if (character == 'a' || character == 'e' || character == 'i' || character
== 'o' || character == 'u') {
return 0;
} else {
return character;
}
}
Your issue (outputting null characters) comes from the fact that you unconditionally putchar(3) the result of remove_all_vowels, which returns 0 (null character) when given character is a vowel.
To replace vowels with spaces:
You can simply change return 0; in remove_all_vowels to return ' ';
To completely remove vowels:
I would suggest having a function just to help you check against vowels rather than having it act like a transformation over a char, which is really pythonest.
Example of code:
int is_vowel(int character) {
return (
character == 'a' || character == 'e' || character == 'i'
|| character == 'o' || character == 'u'
);
}
// Then, in your main...
...
if (!is_vowel(character))
putchar(character);
your program works fine here
https://www.onlinegdb.com/online_c_compiler
Can give a screenshot whats the problem of your compiler?
#include <stdio.h>
#include <string.h>
int check_vowel(char);
int main()
{
char s[100], t[100];
int c, d = 0;
printf("Enter a string to delete vowels\n");
gets(s);
for(c = 0; s[c] != '\0'; c++) {
if(check_vowel(s[c]) == 0) { // If not a vowel
t[d] = s[c];
d++;
}
}
t[d] = '\0';
strcpy(s, t); // We are changing initial string. This is optional.
printf("String after deleting vowels: %s\n", s);
return 0;
}
int check_vowel(char ch)
{
if (ch == 'a' || ch == 'A' || ch == 'e' || ch == 'E' || ch == 'i' || ch == 'I' || ch =='o' || ch=='O' || ch == 'u' || ch == 'U')
return 1;
else
return 0;
}
I am constructing a program that takes string input from the keyboard then shows the number of consonants as an output. I have managed to do it in a ridiculous way in the function count_consonants. I tested using if statement whether each character in the input is a number or symbol to ignore them during calculations. I originally wanted to check if the string is not a string using fgets but I don't know how. That's not an effective way, so any ideas for this?
#include <stdio.h>
#include <string.h>
//function to calculate the consonants
int count_consonants(char str[]) {
int idx;
for (idx = 0; idx < 100; ++idx) {
if (str[idx] == '\0') {
break;
}
}
int vowl = 0;
for (int i = 0; i < idx; ++i) { //loop to check if the characters are vowels or not
if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o'
|| str[i] == 'u' || str[i] == 'A' || str[i] == 'E' || str[i] == 'I'
|| str[i] == 'O' || str[i] == 'U' || str[i] == ' ') {
vowl += 1;
}
// numbers and symbols are counted here as vowels because if not,
// the compiler will count them the other way around
if (str[i] == '1' || str[i] == '2' || str[i] == '3' || str[i] == '4'
|| str[i] == '5' || str[i] == '6' || str[i] == '7' || str[i] == '8'
|| str[i] == '9') {
vowl += 1;
}
if (str[i] == ':' || str[i] == ',' || str[i] == '.' || str[i] == '$'
|| str[i] == '%' || str[i] == '^' || str[i] == '&' || str[i] == '*'
|| str[i] == '#' || str[i] == '_' || str[i] == '!') {
vowl += 1;
}
}
int cons = idx - vowl; // consonants = whole length of text - vowels
return cons - 1;
}
int main(int argc, char const *argv[]) {
char string[100];
char store[100][100];
int i = 0;
while (string[0] != '\n') {
fgets(string, 100, stdin);
strcpy(store[i], string);
i++;
}
for (int j = 0; j < i - 1; ++j) {
/* code */
printf("Number of consonants=%d\n", count_consonants(store[j]));
}
return 0;
}
shows the number of consonants
A simply way to count consonants, use isalpha(), strchr()
#include <string.h>
#include <ctype.h>
int my_isavowel(char ch) {
const char *p = strchr("aeiouAEIOU", ch); // search for a match
return p && *p; // If p is not NULL, and does not point to \0
}
int count_consonants(const char str[]) {
int count = 0;
while (*str != '\0') { // while not at end of string ...
char ch = *str++; // Get character and advance
count += isalpha((unsigned char) ch) && !my_isvowel(ch);
}
return count;
}
If you look for number of consonants, simply best count consonants instead of other things
#include <stdio.h>
#include <string.h>
int main (int narg,char*args[]){
char cons[ ] = "ZRTPQSDFGHJKLMWXCVBN";
char sentence[ ] = "This is my sentence!";
int i=0;
int sum_cons = 0;
for (i = 0; i < strlen(sentence); ++i)
if (strchr(cons,strupr(sentence)[i])) sum_cons++;
printf ("#CONS>%i\n",sum_cons);
return 0;
}
I'm having difficulties with a program I've been instructed to write. The program should search a word for the first vowel that appears, it then should print out the index of that vowel. If there are no vowel's, it should return -1.
This is my code so far:
int firstVowel(char* string){
//Variable for case with no vowels
int notInString = -1;
int length = strlen(string);
int i;
for(i = 0; i <= length; i+=1){
if(*string == 'a' || *string == 'e' || *string == 'i' ||
*string == 'o' || *string == 'u'){
return i;
}
else if(*string != 'a' || *string != 'e' || *string != 'i' || *string != 'o' ||
*string != 'u') {
return notInString;
}
}
}
When I run it with the input "abced" it returns 0 correctly. However, when I run it as fsed it returns -1 incorrectly.
You loop does not increment string and always returns after doing the first comparison, you want
int firstVowel(char* string) {
//Variable for case with no vowels
int notInString = -1;
int length = strlen(string);
int i;
for (i = 0; i <= length; i += 1) {
if (*string == 'a' || *string == 'e' || *string == 'i' ||
*string == 'o' || *string == 'u') {
return i;
} else {
string++;
}
}
return notInString;
}
This will search the entire array, returning the index of the first vowel it sees. It cannot return notInString until it has processed the entire string.