Thread: C programing need help.

You don't need to use calloc() there, malloc() is fine.

For the remove:

1) find the first letter of the word with strstr().
2) replace it with '\0'.
3) advance the pointer returned by strstr by the length of the word.
4) strcat the original str and the pointer together.

The reason for adding '\0' is to terminate the string there, but this does not replace the rest of the data in the array. So when you call strcat(), the data you're pointer points to will be tacked on at the '\0'. You can then realloc smaller by the length of the word.

Replace is the same thing with a couple more steps. So try and get the remove to work first. If you have trouble, post it here.

Also, you should limit the input length this way to prevent a buffer overflow on word:

Code:

scanf("%19s", word);  // max 19 characters

Making word substantially larger -- say 256 bytes -- might be a good idea too. That is still a very trivial amount of memory, and it is on the stack anyway, which has a preset size usually in the MB's.

Originally Posted by MK27

1) find the first letter of the word with strstr().
2) replace it with '\0'.
3) advance the pointer returned by strstr by the length of the word.
4) strcat the original str and the pointer together.

The reason for adding '\0' is to terminate the string there, but this does not replace the rest of the data in the array. So when you call strcat(), the data you're pointer points to will be tacked on at the '\0'. You can then realloc smaller by the length of the word.

Replace is the same thing with a couple more steps. So try and get the remove to work first. If you have trouble, post it here.

how will this help me check whether the word is anagram?

Originally Posted by badblueboy

how will this help me check whether the word is anagram?

It won't, I suppose you would have to do that first, but determining whether a word is an anagram or not should be fairly simple. Make that a separate function, and loop thru the sentence using something like this:

Code:

#include <stdio.h>
#include <string.h>

int main(void) {
	char str[] = "the quick brown fox jumped over the lazy red dog",
		word[128], *p = str;

	while (*p && sscanf(p, "%127s", word)) {
		puts(word);
		p += strlen(word);
		while (*p == ' ') p++;
	}
	return 0;
}

In place of the puts(), you would be checking for an anagram and removing it if necessary. That will require a bit more logic WRT (not) advancing the pointer afterward.

Perhaps I am wrong explain, but I must find anagreme contained in a sentence.
example:
imput:
i love, vole and snake.
output:
love, vole.

Sorry, I was thinking of palindromes, but I guess there aren't many individual words like that.

I guess you would want to start by checking if the words are the same length. Then use strchr() on the second word for each letter from the first word. If you make it thru all the letters, the words are anagrams.

Are there non-ascii characters in Croatian? If you have to work with that, it's going to be trickier.

WRT to punctuation, you might want to write a function to get the length of the word sans trailing punctuation:

Code:

#include <stdio.h>
#include <string.h>

int wordlength (char *word) {
	int i = 0;
	while ( (word[i] >= 'a' && word[i] <= 'z')
	  || (word[i] >= 'A' && word[i] <= 'Z') )
		i++;
	return i;
}

int main(void) {
	char str[] = "the. quick; brown. Fox,",
		word[128], *p = str;

	while (*p && sscanf(p, "%127s", word)) {
		printf("%s %d\n", word, wordlength(word));
		p += strlen(word);
		while (*p == ' ') p++;
	}

	return 0;
}

Output:
the. 3
quick; 5
brown. 5
Fox, 3


Based on the specification in the OP, "Punctuation should be added immediately after the word", so this should work to give you the word length for the purpose of the anagram check.

Only the English alphabet



I would really appreciate if you could write me that code.

Originally Posted by badblueboy

I would really appreciate if you could write me that code.

Nope.
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I've given you quite a bit of code as it is. Go back and see if you can write a function that will check if two words are anagrams, as per my suggestion in post #9. If you have trouble, post your attempt.

because rij (horrible variable name btw) is an array, which gets converted to a pointer in that context. rj[i] is a char, which is an "integer type". So it's trying to convert a char pointer to a char. Also, rj is not pointing to anything, so it will likely crash if you get it to compile.