something like strtok

The problem:
I have a string in a pointer or in an array.I want to seperate the words from this string and drop them into a list.Example: we have string "my name is Bond". I want to seperate the words like: "my" and drop it to the list in the first node, "name" and drop it in the second node, "is" and "Bond" the same.
Forget the list, I know how to make this
My problem is that i must write a fynction to seperate the words in a string but i don't know how to begin.Any idea? thanks!

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What are the characters that separate the words? How would you define a set of instructions to separate the words from each other, that should apply to all sentences?

Wouldn't be something like..

1. Find the first space.
2. Read characters from 0 to first space.
3. Find next space.
4. Read characters from (position of first space + 1) to next space.
5. Rinse and repeat 3 and 4 until finished.

Now program.

And what, exactly, is wrong with using strtok() for this?

--
Mats

@matsp
i must write my function

@IceDane
Yes i thought something like this but i think that:
every time i find a space i must replace it with \0 or not? Because i must pass this word as a string into a list.
Another question: my thought is to use an array of pointers to have the words and then drop its word into the list.Am i correct?

Ah, ok. Are you allowed to use ANY str* functions? Would for example "strchr" or "strspn" be valid to use?

And yes, you have to replace the space with a zero character. [Actually, you don't have to, if you use some other method to remember how long the word was, but I don't see why you would want to use that method - it just complicates matters].

Yes, using an array of pointers, as long as it's long enough - worst case is half the number of characters in the string [e.g. "a a a a a a a a a a"].

--
Mats

I can use only strlen and strcpy from str functions
So you say that is possible to have an array like that : "hello/0my/0name/0is/0"
But how i pass the string to the array from strings( char *str[5]) with strcpy? should i remember everytime where is the first letter from each word example h m n and i ?

which method did you mean that is more simple? this with "strchr" or "strspn"?

No, you still have just one char *, not an array of char *.

If you create a new char *[], then each entry will point to a new string, e.g.

Code:

char *strings[SOMENUMBER]; strings[0] = "my"; strings[1] = "name"; strings[2] = "is";

strcpy() needs to be passed one of those strings at a time.

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Mats

Code:

int len; char t[10]="hello wor"; int i, j , last; char *words[3]; i=0; len=strlen(t); while(t[i]<len){ if(t[i]= ' '){ last=i; strcpy(words[0], t);} i++; } printf(words[0]);

don't seems to work? I can't understand what will be the second argument of strcpy

That code definitely doesn't work since you haven't allocated any memory for "words[0]" - you only have the memory for the words array itself, but there is no memory for the strings in there.

Also, you don't set the space to zero, so the string is still "Hello Wo" when you do the copy.

In the final code, you will also need to have a pointer to where in t you started the search for the second word.

--
Mats

so i need malloc to allocate memory for *words[0] ?
Maybe the idea of array *words[3] is not so good .Is there a way to do this without the use of malloc

some changes

Code:

int len,l; char t[10]="hello wor"; int i, j , last; char words[7]; i=0; len=strlen(t); while(t[i]<len){ if(t[i]= ' '){ last=i; t[i]='\0'; strcpy(words, t);} i++; t[0]=t[i];/*set the t to point to "wor"*/ }

You could set "words[x]" to the beginning of the current word, perhaps.

And you can't move the entire string with

Code:

t[0] = t[i]

, that just sets the first char to the one at t[i] (presumably 'w' - so now t is "wello wo" instead).

--
Mats

Code:

i=0; len=strlen(t); while(i<len){ if(t[i]= ' '){ last=i; t[i]='\0';} printf(t); i++; }

Ok first this part of code.I think that this code should print only hello but it prints hello wor

I'm surprised it prints anything visible at all, actually - you are overwriting everything with space, because you are using a single equal sign when you should be comparing to space using two equal signs.

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Mats

i don't understand. You mean that i should use if(t[i]== ' ') or something else?

Quote:

Originally Posted by alzar

i don't understand. You mean that i should use if(t[i]== ' ') or something else?

Yes, that would prevent "t[i]" from being overwritten with a space in the if-statement.

This doesn't explain why the printout is as it is.

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Mats

i fix it

Code:

char *new; i=0; len=strlen(t); while(i<len && t[i]!='\0'){ if(t[i]== ' '){ last=i; t[i]='\0'; printf(t); } i++; new = &t[i]; }

and now it prints me only hello
can i drop it immediatly on the list know or i should use another array?
and i use pointer new to catch the new adress of the string "wor"
but how will make this for any string? I mean in the while loop in the if statement know i should check for the pointer new if has a space?

Sure, just strcpy() to copy the string. But next you need to move the looking for next space by looking from "t[i]" onwards. So you probably should introduce another char pointer, which starts at the beginning of "t", then moves to "t[i]" when the space has been found.

I would also use a for-loop, rather than a while-loop, since you are just counting up i all the time. Also you need to copy the last string, even if you don't find a space.

--
Mats

i make a try but it doesn't work. I am confused with the loop.

Code:

int len,l; char t[23]="hello world how are you"; int i, last,count; char *words[10]; char *new; len=strlen(t); new=t; for(i=0; i<len; i++) if(new[i]==' '){ t[i]='\0'; len=strlen(t);/*so this changes also the len in the loop*/ words[count]= malloc(len2*sizeof(char)); strcpy(words[count],t); count++; new=&t[i+1]; } }

any help?

I used your program as a template, and came up with this. It's not very elegant, but it's clear what it's doing, I think.

Code:

#include <stdio.h> #include <stdlib.h> #include <string.h> int main(void) { int len,l; char t[]="hello world how are you"; int c, i, last, count; int offset = 0; char *words[10] = {0}; char *new; len=strlen(t); for(i = 0, count = 0; i <= len; i++) if(t[i]==' ' || i == len) { t[i]='\0'; if(offset == 0) words[count] = malloc(i * sizeof(char)); else words[count] = malloc(offset * sizeof(char)); strcpy(words[count], t + offset); count++; offset = i + 1; } for(i = 0; i < count; i++) printf("\n %s", words[i]); printf("\n\n Press Enter When Ready "); c = getchar(); return 0; }

sizeof(char) is always 1, and I have a sneaking suspicion that you're malloc()'ing one less byte than you ought to be . . . .

Oh, and l (lowercase L) is an absolutely stupid name for a variable -- it looks just like a 1.

(@Adak: I realize that most of your code was based on alzar's, but still . . . .)

Quote:

Originally Posted by dwks

sizeof(char) is always 1, and I have a sneaking suspicion that you're malloc()'ing one less byte than you ought to be . . . .

Oh, and l (lowercase L) is an absolutely stupid name for a variable -- it looks just like a 1.

(@Adak: I realize that most of your code was based on alzar's, but still . . . .)

That's funny, I removed several variables, added one's that made sense to me, name-wise, and never once even noticed the lower-case L. I don't use that for a variable name.

The malloc length is right in very limited testing. The size reflects that the cp is either at the space AFTER the word, or at the NULL AFTER the last word. The last NULL gets written over and replaced by another NULL, but it wasn't worth changing the logic for that, imo.

I didn't try that code out on other sentences/words, so I ain't a guaranteeing nuttin'! :D

Student's beware!! :D :D :D

Thank you so match Adak i try and other sentences and it seems to work fine.It was so easy.I was trying with the pointer new and you didin't need to use this pointer.Thanks!