Here I go again...arrays....char...

This is a discussion on Here I go again...arrays....char... within the C Programming forums, part of the General Programming Boards category; My code is here! Vowels from sentence in one array, all none vowels in other array. First one is working ...

  1. #1
    Sometimes so stupid... shardin's Avatar
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    Here I go again...arrays....char...

    My code is here! Vowels from sentence in one array, all none vowels in other array. First one is working but I can't get it to put all others in other array! Help!

    Code:
    int main()
    
    {
    	int sranje;
    	char niz[100];
    	char niz2[100]={NULL};
    	char niz4[100]={NULL};
    	char niz3[]={'a','A','e','E','U','u','i','I','o','O'};
    	int a,b,c=0,d=0,br=0;
    
    	printf("Upisi niz: ");
    	fgets(niz, sizeof(niz), stdin);
    
    for(a=0;a<strlen(niz)+1;a++)
    {
    		if(niz[a]==' ')
    			br++;
    	for(b=0;b<strlen(niz3)+1;b++)
    	{
    		if(niz[a]==niz3[b])
    		{
    			niz2[c]=niz[a];
    			c++;
    		}
    	}
    }
    		puts(niz2);
    		printf("\n");
    		puts(niz4);
    		printf("\n%d", br);
    
    	scanf("%d", &sranje);
    }
    I tried with combinations like this,
    Code:
                               f(niz[a]!=niz3[b])
    			{
    				niz4[d]=niz[a];
    				d++;
    			}
    but, nothing.

  2. #2
    CSharpener vart's Avatar
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    strlen(niz3)

    to use this your niz3 should be null-terminated.
    In your case - it is not. It is just a char array, so no string manipulation routines can be applied.
    Or use sizeof(niz3)
    Or initialize it in another way - to get the null-termination character. For example
    Code:
    char niz3[]="aAeEUuiIoO";
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

  3. #3
    Sometimes so stupid... shardin's Avatar
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    I did and if I implement this

    Code:
    if(niz[a]!=niz3[b])
    			{
    				niz4[d]=niz[a];
    				d++;
    			}
    after first if, and input is "aaa", then it returns for "niz4" - aaaaaaaaaaaaaaaaaaaa

  4. #4
    Algorithm Dissector iMalc's Avatar
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    Adding one to the results of strlen is basically saying "explicit buffer overrun here please". Then leaving off the nul terminator char, well you just really really want to crash something don't you?

    I advise single-stepping the whole way through your program.
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  5. #5
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    Well the a proper solution is to test every letter of the input buffer to see whether is a vowel or not. Each time add every letter to its category and get the final results.
    I think that a simple answer code is the following, think of bugs included.

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    
    const char *Vowls = "aeiou";
    
    int GetVowels(const char *Buf, char **Vowels, char **NoVowels)
    {
    	if(Buf == NULL)
    	{
    		printf("Empty Buffer.\n");
    		return -1;
    	}
    	else
    	{
    		//Variables.
    		int i;
    		int j = 0;
    		int n = 0;
    		//Get memory for the pointers.
    		*Vowels = calloc(1, sizeof(char));
    		*NoVowels = calloc(1, sizeof(char));
    		//Check for memory fault.
    		if(*Vowels == NULL || *NoVowels == NULL)
    		{
    			printf("Memory Error.\n");
    			return -2;
    		}
    		else
    		{
    			//Search each letter seperately to find what you need, save it to char arrays.
    			for(i = 0; i < (int)strlen(Buf); i++)
    			{
    				//Check for valid input, you can do whatever you want with it, i prefer to break as an invalid input.
    				if(Buf[i] == ';' || Buf[i] == ',' || Buf[i] == '!' || Buf[i] == ':')
    					break;
    				if(strchr(Vowls, Buf[i]) != NULL)
    				{
    					Vowels[0] = realloc(Vowels[0], sizeof(char) * (j+1));
    					*(Vowels[0]+ j) = Buf[i];
    					j++;
    				}
    				else
    				{
    					NoVowels[0] = realloc(NoVowels[0], sizeof(char) * (n+1));
    					*(NoVowels[0]+ n) = Buf[i];
    					n++;
    				}
    			}
    			if(i == strlen(Buf))
    			{
    				//Get space for the '\0'.
    				Vowels[0] = realloc(Vowels[0], sizeof(char) * (j+1));
    				NoVowels[0] = realloc(NoVowels[0], sizeof(char) * (n+1));
    				//Put it.
    				*(Vowels[0]+ j) = '\0';
    				*(NoVowels[0]+ n) = '\0';
    				return 0;
    			}
    			else
    			{
    				printf("Internal Loop Error, wrong misplaced symbol &#37;c in Buffer in place %d.\n", Buf[i], i);
    				free(*Vowels);
    				free(*NoVowels);
    				*Vowels = NULL;
    				*NoVowels = NULL;
    				return -3;
    			}
    		}
    	}
    }
    int main(int argc, char *argv[])
    {
    	char Buf[256] = "";
    	char *nv = NULL;
    	char *v = NULL;
    	printf("Give me the sentance:");
    	fgets(Buf, sizeof(Buf), stdin);
    	if(Buf[strlen(Buf) - 1] == '\n')
    		Buf[strlen(Buf) - 1] = '\0';
    	GetVowels(Buf, &v, &nv);
    	printf("NoVowels: %s\n",nv);
    	printf("Vowels:   %s\n",v);
    	free(v);
    	free(nv);
    	return 0;
    }
    I dont remember exactly if the Vowels are these i declared but you can easily add or remove letters from the string before the function!!!

    Some results:

    Give me the sentance:Hi this is the new way of making code less friendly
    NoVowels: H ths s th nw w f mkng cd lss frndl
    Vowels: iiieeayoaioeeiey
    Press any key to continue . . .

    Bokarinho!
    Last edited by Bokarinho; 09-10-2007 at 02:37 AM.

  6. #6
    Registered User whiteflags's Avatar
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    Bokarinho, that's an awful lot of energy to expend on a simple sieve. You sure like to be verbose! There exists a function in the standard C string library that can make such things a trivial exercise.

  7. #7
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by citizen View Post
    Bokarinho, that's an awful lot of energy to expend on a simple sieve. You sure like to be verbose! There exists a function in the standard C string library that can make such things a trivial exercise.
    Seconded. That's a wicked amount of wheel reinvention there. That and 'y' is not a vowel.
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  8. #8
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    Quote Originally Posted by iMalc View Post
    Seconded. That's a wicked amount of wheel reinvention there. That and 'y' is not a vowel.
    Thanks for your great commentary.... Yes y is not a vowel you can easily remove it from the string and its all ok!!! Whats the problem with you people??????

  9. #9
    Technical Lead QuantumPete's Avatar
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    Programmers are pedantic by nature

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  10. #10
    Sometimes so stupid... shardin's Avatar
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    Bokarinho
    I think that a simple answer code is the following, think of bugs included.
    thx, but i think there is a much simpler way to do this, but i can learn something from your code, if i can't use that in this task ill use it in another.

  11. #11
    Sometimes so stupid... shardin's Avatar
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    Adding one to the results of strlen is basically saying "explicit buffer overrun here please". Then leaving off the nul terminator char, well you just really really want to crash something don't you?
    can you please be more specific. My new code looks like this now:
    Code:
    	char niz[100];
    	char niz2[100]={NULL};
    	char niz4[100]={NULL};
    	char niz3[]="aAeEiIoOuU";
    	int a,b,c=0,d=0,br=0;
    	int i, j, temp=0;
    
    	printf("Upisi niz: ");
    	fgets(niz, sizeof(niz), stdin);
    
    for(a=0;a<strlen(niz);a++)
    {
    		if(niz[a]==' ')
    		{
    			br++;
    		}
    	for(b=0;b<strlen(niz3);b++)
    	{
    		if(niz[a]==niz3[b])
    		{
    			niz2[c]=niz[a];
    			c++;
    		}
    		if(niz[a]!=niz3[b])
    		{
    			niz4[d]=niz[a];
    			d++;
    		}
    		 
    	}
    }
    like i said before, vowels ok aa , novowels puts : aaaaaaaaaaaaaaaaaaaaaaaa

  12. #12
    Registered User whiteflags's Avatar
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    > like i said before, vowels ok aa , novowels puts : aaaaaaaaaaaaaaaaaaaaaaaa

    You need to work ouy a way to check all of the vowels at once, given some letter in an input string. Right now, you're code only looks for 'a'. What you have simply doesn't work for the same reason that 'a' is not 'e' or any other letter.

    Look up a function called strchr(), find out what it does, and think about how it can help you with your new problem. Or strcspn(), again.

  13. #13
    Sometimes so stupid... shardin's Avatar
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    Well, fu**'it, I can always do it like this, nice and sorted!!!

    Code:
    #include <stdio.h>
    #include <string.h>
    int main()
    {
    	int sranje;
    	char niz[100];
    	char niz2[100]={NULL};
    	char niz4[100]={NULL};
    	char niz3[]="aAeEiIoOuU";
    	char niz5[]="bBcCdDfFgGhHkKlLpPwWrRtTzZQqsSyYxXjJVMmnN";
    	int a,b,c=0,d=0,br=0;
    	int i, j, temp=0;
    
    	printf("Upisi niz: ");
    	fgets(niz, sizeof(niz), stdin);
    
    for(a=0;a<strlen(niz)-1;a++)
    {
    		if(niz[a]==' ')
    		{
    			br++;
    		}
    	for(b=0;b<strlen(niz3)-1;b++)
    	{
    		if(niz[a]==niz3[b])
    		{
    			niz2[c]=niz[a];
    			c++;
    		}
    	}
    	for(b=0;b<strlen(niz5)-1;b++)
    	{
    		if(niz[a]==niz5[b])
    		{
    			niz4[d]=niz[a];
    			d++;
    		}
    	}
    }
    for(c=0;c<strlen(niz2);c++)
    {
    	for(j=0;j<strlen(niz2);j++)
    	{
    		if(niz2[c]<niz2[j])
    		{
    			temp=niz2[c];
    			niz2[c]=niz2[j];
    			niz2[j]=temp;
    		}
    	}
    }
    for(c=0;c<strlen(niz4);c++)
    {
    	for(j=0;j<strlen(niz4);j++)
    	{
    		if(niz4[c]<niz4[j])
    		{
    			temp=niz4[c];
    			niz4[c]=niz4[j];
    			niz4[j]=temp;
    		}
    	}
    }
    		puts(niz2);
    		printf("\n");
    		puts(niz4);
    		printf("\n&#37;d", br);
    
    	scanf("%d", &sranje);
    }
    Last edited by shardin; 09-10-2007 at 04:14 AM.

  14. #14
    Technical Lead QuantumPete's Avatar
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    You forgot jJ in your consonant string...
    and you've got cC in there twice...

    QuantumPete
    "No-one else has reported this problem, you're either crazy or a liar" - Dogbert Technical Support
    "Have you tried turning it off and on again?" - The IT Crowd

  15. #15
    Sometimes so stupid... shardin's Avatar
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    it's there now...
    ...and aprentice shall become master...or not...

    "Never let your sense of moral prevent you from doing what is right!" Salvor Hardin, mayor of Terminus

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