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Can you write a c program to count the number of words with not more than 5 consonants in a text file?
Wiki User
∙ 11y ago
FILE *fPtr = fopen("ConsonantInput.txt", "r");
char sWord[1024];
int nWordCount = 0;
while(!feof(fPtr))
{
memset(sWord, 0x00, sizeof(sWord));
fscanf(fPtr, "%s", sWord);
short nCount = strlen(sWord);
short nConsonantCount = 0;
for(int i = 0; i < nCount; ++i)
{
switch(sWord[1])
{
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
case '-':
{
break;
}
default:
++nConsonantCount;
}
if(nConsonantCount >= MAX_CONSONANT_NUMBER)
{
++nWordCount;
break;
}
}
}
printf("Number of Words : %d, Time Taken : %.2f", nWordCount);
fclose(fPtr);
Wiki User
∙ 11y ago
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Program to count the number of characters and words in the line?
in linux wc -l filename will count the lines and wc will count the letters
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consonants with BB
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Words that have consonants in them are most commonly the words that do not have all vowels. Some examples of such words are "cat," "dog," "book," and "can."
How many words with or without meaning each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants namely, D, G, H, T, and R. Number of ways of selecting 2 vowels out of 3 vowels = Number of ways of selecting 3 consonants out of 5 consonants = Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30 Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! ways. Hence, required number of different words = 30 × 5! = 3600
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Please refer to the related questions for some words start and end with consonants.
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What number words have two vowels and three constanants?
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What is summing the number of words in a document?
Word Count
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You have written 6 of them in your question - what, words, consonants, next, each, other I hope this helps
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alot
