I am trying to write a program that adds, subtracts, multiplies, and divides a string of characters. Where I'm at now with the program is figuring out how to split the input string into two strings, and then perform the appropriate +-/*.
The input should look like this abc+aaa
and the output for that should be abc + aaa = bcd
How do I convert character strings into integer strings?
#include <stdio.h>
#include <math.h>
#include <string.h>
int main() {
printf("This is a pseudo arithmetic program");
char input[10];
input[10] = '\0';
char first [9];
first[9] = '\0';
char last [9];
last[9] = '\0';
int i = 0;
int b;
int e;
while (input[0] != '0') {
if (input[0] == 0){
return -1;
}
printf("\nEnter a math problem in SOS format using only lowercase letters up to 9 characters");
printf("\nEx: abc+abc... type '0' to quit \n");
scanf("%s", input);
int x = 0;
x = strlen(input);
if (strchr(input, '+')){
for (i = 0; i <= x; i++) {
if (i == '+')
strncpy(first, &input[0], i-1);
i = 0;
}
for (i = x; i >= input[0]; i--) {
if (i == '+')
strncpy(last, &input[i], x);
i = 0;
}
printf("%s", first);
printf(" + ");
printf("%s", last);
printf(" = %d", first + last);
}
There seems to be multiple problems with your code:
There is a array out of bounds happening for almost all the arrays:
char input[10];
input[10] = '\0';
In this if you want to initialize the last character with '\0' then it should be
input [9] = '\0'
Arrays indexes always start from 0.
It is not clear what is the use of below lines:
while (input[0] != '0') { if (input[0] == 0){ return -1; }
When taking input for a string, why are prompting users to enter a 0 to end it?
strrchr returns the pointer from where the searched character begins. So, you can that itself to determine where the '+' symbol is and two split the strings instead of your while loop. See strrchr man page
Also, your idea of adding characters is not clear. From your example, it appears you are considering a = 1, b = 2 etc. In such a case, if your code is case insensitive, then you can convert all your input to upper case and then do (input[0] - 'A')+1 to convert your letters like a, b, c to 1, 2, 3 etc.
Hope these pointers help. Suggest you check your problem statement again and refactor your code accordingly.
Related
I'm writing a code that must identify the letter 't' or 'T' in a word, before or after the middle of it.
If the first half of the word does contain a 't' or a 'T', the program should output a 1. If the first half does not contain the letter 't' or 'T', but the second half does, then the program should output a 2. Otherwise, if there is no 't' or 'T' in the word at all, the program's output should be -1. The word entered will not have more than 50 letters.
#include <stdio.h>
#include <string.h>
int main() {
char word[50];
int i = 0, length, t = 0, T = 0;
scanf("%s", word);
length = strlen(word);
t = word[i] == 't';
T = word[i] == 'T';
while(!t || !T) {
if((t || T) && i <= length / 2) {
printf("%d", '1');
} else if((t || T) && i > length / 2) {
printf("%d", '2');
//}else{
// printf("%d", '-1');
}
i++;
}
return 0;
}
If I enter any word and press enter, nothing is printed. Another thing is that when I remove the comment slashes from the two lines at the bottom, the program goes through an infinite loop.
Could someone please help?
This sounds like a school assignment, so I'll focus on advising/critiquing your code rather than giving a solution.
The first recommendation I have is to use a for loop instead of a while loop. A Rule of thumb in C is to only use a while loop when you actually don't have any idea how many things you need your program to look at.
You already have the length of the string, so set up your for loop to loop exactly once for each character.
Next you need to change how you are using printf. The %d format specifier is for printing integers, but you are passing it '1'. This is not an integer, it is the ascii representation of the symbol 1 (which is actually has the value 49, see the ascii table for more info)
You can either pass printf the value 1, or use the %c specifier, which expects ascii characters.
Better yet, just say printf("1");
That doesn't get you all the way there, but I think it lays the ground work so you can find the solution!
Condition !t || !T has no sense to be used as loop condition ...ask yourself how the loop will end ? you need just to check i is less than length
Second, the assignments t = word[i] == 't'; T = word[i] == 'T'; outside the loop have no sense ...you will be just pointing to the zero index of the string ...you should check all characters
third , the printf lines need to use %d
fourth , you appear not getting the purpose of the program printing inside loop will lead to printing many numbers and you just want to know if there is t or T you need to print single line.you may use variable int result=0; to hold the value you want and print it in the end ...of course you will need using break statement in the if((t || T) && i <= length / 2) and if((t || T) && i > length / 2) because no need for more searching
fifth, you should re-read , re-think , re-code the assignment before going bored and asking about it
sixth, there is a working version by modifying your code but you should try writing a good solution before looking at a solution as it better to solve your problems by yourself
#include <stdio.h>
#include <string.h>
int main() {
char word[50];
int i = 0, length, t = 0, T = 0;
scanf("%s", word);
length = strlen(word);
int result=0;
while( i<length) {
t = word[i] == 't';
T = word[i] == 'T';
if((t || T) && i <= length / 2) {
result=1;
break;
} else if((t || T) && i > length / 2) {
result=2;
break;
}else{
result=-1;
}
i++;
}
printf("%d",result);
return 0;
}
# include <stdio.h>
int main()
{
char name[20];
int age;
int siblings;
int childrens;
printf ("Hello my name is A.I, what is your name? \n");
scanf("%s", &name);
printf("how old are you : \n");
scanf("%d",&age);
printf("how many siblings you have: \n");
scanf("%d", &siblings);
printf("how many children you have: \n");
scanf("%d", &childrens);
printf("so your name is : %s \n", name);
printf("and your age is : %d \n", age);
printf("you have siblings : %d\n", siblings);
printf("so you have childrens : %d\n", childrens);
return 0;
}
This is for Homework
I have to write a program that asks the user to enter a string, then my program would separate the even and odd values from the entered string. Here is my program.
#include <stdio.h>
#include <string.h>
int main(void) {
char *str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
scanf("%s", &str);
while (&str[i] < 41) {
if (i % 2 == 0) {
odd[j++] = *str[i];
} else {
even[k++] = *str[i];
}
i++;
}
printf("The even string is:%s\n ", even);
printf("The odd string is:%s\n ", odd);
return 0;
}
When I try and compile my program I get two warnings:
For my scanf I get "format '%s' expects arguments of type char but argument has 'char * (*)[41]". I'm not sure what this means but I assume it's because of the array initialization.
On the while loop it gives me the warning that says comparison between pointer and integer. I'm not sure what that means either and I thought it was legal in C to make that comparison.
When I compile the program, I get random characters for both the even and odd string.
Any help would be appreciated!
this declaration is wrong:
char *str[41];
you're declaring 41 uninitialized strings. You want:
char str[41];
then, scanf("%40s" , str);, no & and limit the input size (safety)
then the loop (where your while (str[i]<41) is wrong, it probably ends at once since letters start at 65 (ascii code for "A"). You wanted to test i against 41 but test str[i] against \0 instead, else you get all the garbage after nul-termination char in one of odd or even strings if the string is not exactly 40 bytes long)
while (str[i]) {
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
if you want to use a pointer (assignement requirement), just define str as before:
char str[41];
scan the input value on it as indicated above, then point on it:
char *p = str;
And now that you defined a pointer on a buffer, if you're required to use deference instead of index access you can do:
while (*p) { // test end of string termination
if (i % 2 == 0) { // if ((p-str) % 2 == 0) { would allow to get rid of i
odd[j++] = *p;
} else {
even[k++] = *p;
}
p++;
i++;
}
(we have to increase i for the even/odd test, or we would have to test p-str evenness)
aaaand last classical mistake (thanks to last-minute comments), even & odd aren't null terminated so the risk of getting garbage at the end when printing them, you need:
even[k] = odd[j] = '\0';
(as another answer states, check the concept of even & odd, the expected result may be the other way round)
There are multiple problems in your code:
You define an array of pointers char *str[41], not an array of char.
You should pass the array to scanf instead of its address: When passed to a function, an array decays into a pointer to its first element.
You should limit the number of characters read by scanf.
You should iterate until the end of the string, not on all elements of the array, especially with (&str[i] < 41) that compares the address of the ith element with the value 41, which is meaningless. The end of the string is the null terminator which can be tested with (str[i] != '\0').
You should read the characters from str with str[i].
You should null terminate the even and odd arrays.
Here is a modified version:
#include <stdio.h>
int main(void) {
char str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
if (scanf("%40s", str) != 1)
return 1;
while (str[i] != '\0') {
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
odd[j] = even[k] = '\0';
printf("The even string is: %s\n", even);
printf("The odd string is: %s\n", odd);
return 0;
}
Note that your interpretation of even and odd characters assumes 1-based offsets, ie: the first character is an odd character. This is not consistent with the C approach where an even characters would be interpreted as having en even offset from the beginning of the string, starting at 0.
Many answers all ready point out the original code`s problems.
Below are some ideas to reduce memory usage as the 2 arrays odd[], even[] are not needed.
As the "even" characters are seen, print them out.
As the "odd" characters are seen, move them to the first part of the array.
Alternative print: If code used "%.*s", the array does not need a null character termination.
#include <stdio.h>
#include <string.h>
int main(void) {
char str[41];
printf("Enter a string (40 characters maximum): ");
fflush(stdout);
if (scanf("%40s", str) == 1) {
int i;
printf("The even string is:");
for (i = 0; str[i]; i++) {
if (i % 2 == 0) {
str[i / 2] = str[i]; // copy character to an earlier part of `str[]`
} else {
putchar(str[i]);
}
}
printf("\n");
printf("The odd string is:%.*s\n ", (i + 1) / 2, str);
}
return 0;
}
or simply
printf("The even string is:");
for (int i = 0; str[i]; i++) {
if (i % 2 != 0) {
putchar(str[i]);
}
}
printf("\n");
printf("The odd string is:");
for (int i = 0; str[i]; i++) {
if (i % 2 == 0) {
putchar(str[i]);
}
}
printf("\n");
here is your solution :)
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[41];
char odd[21];
char even[21];
int i = 0;
int j = 0;
int k = 0;
printf("Enter a string (40 characters maximum): ");
scanf("%s" , str);
while (i < strlen(str))
{
if (i % 2 == 0) {
odd[j++] = str[i];
} else {
even[k++] = str[i];
}
i++;
}
odd[j] = '\0';
even[k] = '\0';
printf("The even string is:%s\n " , even);
printf("The odd string is:%s\n " , odd);
return 0;
}
solved the mistake in the declaration, the scanning string value, condition of the while loop and assignment of element of array. :)
This question already has answers here:
How do I check if a number is a palindrome?
(53 answers)
Closed 5 years ago.
I am trying to check if an input number is a palindrome. I am doing it through strings rather than ints. So, I am taking in a string and reversing it into another string. However, when I use the string compare function it does not give me 0, stating that the strings are not the same. Even when I put in for example "1001", both the input and reverse strings displays 1001. I have figured it out with other methods but am trying to understand what is wrong with this one in specific.
#include <stdio.h>
#include <string.h>
int main(void)
{
char input[100];
char reverse[100];
int numLen = 0;
printf("Enter a number\n");
fgets(input, 100, stdin);
printf("The number is: %s\n", input);
numLen = strlen(input) - 1;
printf("Length of string is: %d\n", numLen);
for (int i = 0; i < numLen; i++)
{
reverse[i] = input[numLen - 1 - i];
if (i == numLen - 1)
{
reverse[i + 1] = '\0';
}
}
printf("The reverse number is: %s\n", reverse);
printf("The original number is: %s\n", input);
int result = strcmp(input, reverse);
printf("Result of strcmp gives us: %d\n", result);
if (strcmp(input, reverse) == 0)
{
printf("These numbers are palindromes\n");
}
else
{
printf("These numbers are not palindromes\n");
}
return 0;
}
The problem is you are not handling the strings properly. You should overwrite the '\n' with \0.
...
char input[100];
char reverse[100];
int numLen = 0;
printf("Enter a number\n");
fgets(input, 100, stdin);
printf("The number is: %s\n", input);
input[strcspn(input,"\n")]='\0'; // getting the length of the
// string without `\n`
// and overwriting with `\0`
numLen = strlen(input) ; // now you don't need to put the -1
printf("Length of string is: %d\n", numLen);
for (int i = 0; i < numLen; i++)
{
....
Apart from these two changes everything else remains the same. You were reversing it all right. And then you used strcmp right way. But the extra \n is removed in the code I have shown.
(still) Why it works?
Now to give you a better idea. You formed the reversed string alright. But the original string has \n within itself.
printf("The reverse number is: (%s)\n", reverse);
printf("The original number is: (%s)\n", input);
In the previous program you just do write these two lines. You will understand where you went wrong.
On giving input 1001Enter it gives this output.
The reverse number is: (1001)
The original number is: (1001
)
What is strcspn doing?
I have using strcspn function got the length without \n and overwriting it with \0.
0 1 2 3 4 5 --> indices
1 0 0 1 \n \0 --> strcspn(input,"\n") returns 4.
1 0 0 1 \0 \0 --> input[strcspn(input,"\n")]='\0'
You can do simply like this without the copying and everything.
Without extra memory - in place palindrome checking
bool checkPal(const char *s){
for(int i = 0, j= strlen(s)-1; i< strlen(s) && j>=0 ; i++)
if(s[i] != s[j])
return false;
return true;
}
int main(void)
{
char input[100];
char reverse[100];
printf("Enter a number\n");
if( fgets(input, 100, stdin) )
printf("The number is: %s\n", input);
input[strcspn(input,"\n")]='\0';
int numLen = strlen(input) ;
printf("Length of string is: %d \n", numLen);
printf("These numbers are %spalindromes\n", checkPal(input)?"not ":"");
return 0;
}
A more succinct way to write the checkPal() would be,
bool checkPal(const char *first){
const char *last = first + strlen(first);
while (first < last) {
if (*first++ != *--last) {
return false;
}
}
return true;
}
last points to the \0 character. Subtraction is necessary before we start doing comparison. To get a clear idea of what happens you have to know the precedence and few rules.
The first<last part is obvious. We are comparing till we reach a point where we first > last (For even length strings) or first = last (for odd length strings).
The if is a bit tricky. *first++ there are two operators involved. * (indirection) and ++(post increment).
And precedence of ++ is higher than de-reference *.
So *first++ will be - first is incremented. Then you might think that we are missing one character very first time but that's not the case. Value of a postfix expression is the value before we do first++. So now you have the first character.
Same way *--last will have the same effect except the value of the prefix expression is the value after the operation. So you are considering the last character.
If they matches we continue. first and last already contain the modified value. We repeat the same logic for rest of the characters in the smaller sub-string.
If a mismatch occurs then we return immediately. (Because it's not a palindrome).
Sorry, my bad. Try this:
#include <stdio.h>
#include <string.h>
// A function to check if a string str is palindrome
void isPalindrome(char str[])
{
// Start from leftmost and rightmost corners of str
int l = 0;
int h = strlen(str) - 1;
// Keep comparing characters while they are same
while (h > l)
{
if (str[l++] != str[h--])
{
printf("%s is Not Palindromen", str);
return;
}
}
printf("%s is palindromen", str);
}
// Driver program to test above function
int main()
{
isPalindrome("abba");
isPalindrome("abbccbba");
isPalindrome("geeks");
return 0;
}
Does this one work?
A variant, recursive version that has no more that the string as argument (or a copy of the original string)
int pal(char *s) {
int n = strlen(s);
if (n <= 1) return 1;
if (s[0] != s[n-1]) return 0;
s[n-1] = '\0';
return pal(++s);
}
return 0: not a palindrome, 1: is a palindrome
Note the string is altered, so you can call it this way if it's a problem (or if the string is created in a static area)
char *copy = malloc(strlen(string)+1); // string is original string
strcpy(copy, string);
int ispal = pal( copy );
printf("Is %s a palindrome\n", ispal ? "":"not");
Reposting because my first post was no good. I have a question that I'm not really sure how to do. I know the process I'm going for, but am not totally sure how to scan a string into an array so that each character/integer is scanned into a independent element of the array. I'll post the question and the code I have so far, and any help would be appreciated.
Question:
Assume that we have a pattern like the following: ([n][letter])+ in which n is an integer number and letter is one of the lowercase letters from a-z. For example, 2a and 3b are valid expressions based on our pattern. Also, “+” at the end of the pattern means that we have at least one expression (string) or more than one expression attached. For instance, 2a4b is another valid expression which is matched with the pattern. In this question, we want to convert these valid expressions to a string in which letters are repeated n times.
o Read an expression (string) from user and print the converted version of the expression in the output.
o Check if input expression is valid. For example, 2ab is not a valid expression. If the expression is not valid, print “Invalid” in the output and ask user to enteranother expression.
o Sample input1 = “2a”, output = aa
o Sample input2 = “2a3b”, output = aabbb
o You will receive extra credit if you briefly explain what concept or theory you can use to check whether an expression is valid or not.
What I have so far:
#include <stdio.h>
int main()
{
int size, i, j;
char pattern[20];
char vowel[20];
int count[20];
printf("Please enter your string: ");
gets(pattern);
size = strlen(pattern);
for(i=0; i<size; i++)
if((i+1)%2 == 0)
vowel[i] = pattern[i];
else if((i+1)%2 != 0)
count[i] = pattern[i];
for(i=0; i<size/2; i++);
for(j=0; j<count[i]; j++)
printf("%s", vowel[i]);
}
I assumed you want to write the "invalid\n" string on stderr. If not just change the file descriptor given to write.
#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <string.h>
#define MAX_INPUT_SIZE 20
int
check_input(char *input)
{
while (*input)
{
if (*input < '0' || *input > '9')
{
write(2, "invalid\n", 8);
return 1;
}
while (*input >= '0' && *input <= '9')
input++;
if (*input < 'a' || *input > 'z')
{
write(2, "invalid\n", 8);
return 1;
}
input++;
}
return 0;
}
void
print_output(char *input)
{
int i;
while (*input)
{
i = atoi(input);
while (*input >= '0' && *input <= '9')
input++;
for (; i > 0; i--)
write(1, input, 1);
input++;
}
write(1, "\n", 1);
}
int
main()
{
char input[MAX_INPUT_SIZE];
do
{
printf("Please enter your string: ");
fgets(input, MAX_INPUT_SIZE, stdin);
input[strlen(input) - 1] = '\0';
}
while (check_input(input));
print_output(input);
return 0;
}
The steps are:
Read pattern
Check if pattern is valid
Generate output
Since the input length is not specified you have to assume a maximum length.
Another assumption is n is a single digit number.
Now you may read the whole expression with fgets() or read it char by char.
The latter allows you to check for validity as you read.
Lets use fgets() for convenience and in case the expression needs to be stored for later use.
char exp[100]; // assuming at most 50 instances of ([n][letter])
int len;
printf("Input: ");
fgets(exp, 100, stdin);
len = strlen(exp) - 1; // Discard newline at end
An empty input is invalid. Also a valid expression length should be even.
if (len == 0 || len%2 != 0) {
printf("Invalid-len\n");
return 1;
}
Now parse the expression and separately store numbers and letters in two arrays.
char nums[50], letters[50];
invalid = 0;
for (i = 0, j = 0; i < len; i += 2, j++) {
if (exp[i] >= '1' && exp[i] <= '9') {
nums[j] = exp[i] - '0';
} else {
invalid = 1;
break;
}
if (exp[i+1] >= 'a' && exp[i+1] <= 'z') {
letters[j] = exp[i+1];
} else {
invalid = 1;
break;
}
}
Notice that in each iteration if first char is not a number or second char is not a letter, then the expression is considered to be invalid.
If the expression is found to be invalid, nothing to do.
if (invalid) {
printf("Invalid\n");
return 1;
}
For a valid expression run nested loops to print the output.
The outer loop iterates for each ([n][letter]) pattern.
The inner loop prints n times the letter.
printf("Output: ");
for (i = 0; i < len/2; i++) {
for (j = 0; j < nums[i]; j++)
printf("%c", letters[i]);
}
This is a rather naive way to solve problems of this type. It is better to use regular expressions.
C standard library doesn't have regex support. However on Unix-like systems you can use POSIX regular expressions.
like this
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>
#define prompt "Please enter your string: "
void occurs_error(const char *src, const char *curr){
printf("\nInvalid\n");
printf("%s\n", src);
while(src++ != curr){
putchar(' ');
}
printf("^\n");
}
bool invalid(char *pattern){
char *p = pattern;
while(*p){
if(!isdigit((unsigned char)*p)){//no number
occurs_error(pattern, p);
break;
}
strtoul(p, &p, 10);
if(!*p || !islower((unsigned char)*p)){//no character or not lowercase
occurs_error(pattern, p);
break;
}
++p;
}
return *p;
}
int main(void){
char pattern[20];
while(fputs(prompt, stdout), fflush(stdout), fgets(pattern, sizeof pattern, stdin)){
pattern[strcspn(pattern, "\n")] = 0;//chomp newline
char *p = pattern;
if(invalid(p)){
continue;
}
while(*p){
int n = strtoul(p, &p, 10);
while(n--)
putchar(*p);
++p;
}
puts("");
}
}
I was wondering if I could ask for some help. I am writing a program in C that writes out the number of characters, words, and vowels are in the string(with a few added print statements). I am trying to figure out how to write a code that loops through the string and counts the number of words that contain at least 3 vowels. I feel as if this is a very easy code to write, but it's always the easiest things that seem to elude me. Any help?
Also: Being new to C, how can I get the same results while using the function int vowel_count(char my_sen[]) instead of using the code within my main?
If that's a tad confusing I mean since my main already contains code to count the number of vowels within my input, how can I somewhat transfer said code into this function and still call upon it in main?
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define SENTENCE 256
int main(void){
char my_sen[SENTENCE], *s; //String that containts at most 256 as well as a pointer
int words = 1, count = 0,vowel_word = 0; //Integer variables being defined
int i,vowel = 0, length; //More definitions
printf("Enter a sentence: ");//Input sentence
gets(my_sen);//Receives and processes input
length = strlen(my_sen); //Stores the length of the input within length
for(i=0;my_sen[i] != '\0'; i++){
if(my_sen[i]=='a' || my_sen[i]=='e' || my_sen[i]=='i' || my_sen[i]=='o' || my_sen[i]=='u' || //Loop that states if the input contains any of the following
my_sen[i]=='A' || my_sen[i]=='E' || my_sen[i]=='I' || my_sen[i]=='O' || my_sen[i]=='U') //characters(in this case, vowels), then it shall be
{ //stored to be later printed
vowel++;
}
if(my_sen[i]==' ' || my_sen[i]=='!' || my_sen[i]=='.' || my_sen[i]==',' || my_sen[i]==';' || //Similar to the vowel loop, but this time
my_sen[i]=='?') //if the following characters are scanned within the input
{ //then the length of the characters within the input is
length--; //subtracted
}
}
for(s = my_sen; *s != '\0'; s++){ //Loop that stores the number of words typed after
if(*s == ' '){ //each following space
count++;
}
}
printf("The sentence entered is %u characters long.\n", length); //Simply prints the number of characters within the input
printf("Number of words in the sentence: %d\n", count + 1); // Adding 1 to t[he count to keep track of the last word
printf("Average length of a word in the input: %d\n", length/count);//Prints the average length of words in the input
printf("Total Number of Vowels: %d\n", vowel);//Prints the number of vowels in the input
printf("Average number of vowels: %d\n", vowel/count);//Prints the average number of vowels within the input
printf("Number of words that contain at least 3 vowels: %d\n", vowel_word);//Prints number of words that contain at least 3 vowels
return 0;
}
It's not much of a problem.
#include <ctype.h>
#include <stdio.h>
#include <string.h>
int vowel_count(char my_sen[])
{
int wcount = 0; // number of words with 3+ vowel chars
int vcount = 0; // current number of vowel chars in the current word
int i = 0; // index into the string
int ch;
while ((ch = my_sen[i++]) != '\0')
{
if (isspace(ch) || !isalpha(ch))
{
// ch is not an alphabetical char, which can happen either
// before a word or after a word.
// If it's after a word, the running vowel count can be >= 3
// and we need to count this word in.
wcount += vcount >= 3; // add 1 to wcount if vcount >= 3
vcount = 0; // reset the running vowel counter
continue; // skip spaces and non-alphabetical chars
}
if (strchr("aeiouAEIOU", ch) != NULL) // if ch is one of these
{
++vcount; // count vowels
}
}
// If my_sen[] ends with an alphabetical char,
// which belongs to the last word, we haven't yet
// had a chance to process its vcount. We only
// do that in the above code when seeing a non-
// alphabetical char following a word, but the
// loop body doesn't execute for the final ch='\0'.
wcount += vcount >= 3; // add 1 to wcount if vcount >= 3
return wcount;
}
int main(void)
{
char sen[] = "CONSTITUTION: We the People of the United States...";
printf("# of words with 3+ vowels in \"%s\" is %d", sen, vowel_count(sen));
return 0;
}
Output (ideone):
# of words with 3+ vowels in "CONSTITUTION: We the People of the United States..." is 3
Btw, you can alter this function to count all things you need. It already finds where words begin and end and so, simple word counting is easy to implement. And word length, too. And so on.
1) Get the string,
2) use strtok () to get each words seperated by space.
3) Loop through each string by char by char to check if it is vowel.
Please check below code
#include<stdio.h>
#include <string.h>
int count_vowels(char []);
int check_vowel(char);
main()
{
char array[100];
printf("Enter a string\n");
gets(array);
char seps[] = " ";
char* token;
int input[5];
int i = 0;
int c = 0;
int count = 0;
token = strtok (array, seps);
while (token != NULL)
{
c = 0;
c = count_vowels(token);
if (c >= 3) {
count++;
}
token = strtok (NULL, seps);
}
printf("Number of words that contain atleast 3 vowels : %d\n", count);
return 0;
}
int count_vowels(char a[])
{
int count = 0, c = 0, flag;
char d;
do
{
d = a[c];
flag = check_vowel(d);
if ( flag == 1 )
count++;
c++;
}while( d != '\0' );
return count;
}
int check_vowel(char a)
{
if ( a >= 'A' && a <= 'Z' )
a = a + 'a' - 'A'; /* Converting to lower case */
if ( a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')
return 1;
return 0;
}