Dynamic strings in C?

Hi guys

One of the things about C that I have also been a little worried about is when using strings. Say that I want to ask the user to input a sentence:

Code:

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#include <stdio.h>

#define length 100

int main()
{
        int i=0;
        char str[length];
       
        while((i < length-1) && (str[i]=getche())!=13)
        {
                i++;
        }

        str[i]='\0';
       
        return 0;
}

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Perhaps the user only enters a sentence 2 characters long. Now, aren't the other 98 characters just taking up unnecessary space?

What is the closest one can get to dynamically allocate a string in C, such that this waste of memory is not present in a program?

Best,
Niles.

Quote:

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Originally Posted by Niels_M

Perhaps the user only enters a sentence 2 characters long. Now, aren't the other 98 characters just taking up unnecessary space?

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Yes, though this might not matter.

Quote:

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Originally Posted by Niels_M

What is the closest one can get to dynamically allocate a string in C, such that this waste of memory is not present in a program?

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One approach is to use malloc() and/or realloc(), expand the dynamic array as needed, and after one is done with reading, trim it down to the required size. Of course, you should also free() appropriately.

By the way, getche is non-standard and it would be better to use character literals instead of magic numbers as such 13.

Quote:

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Originally Posted by laserlight

One approach is to use malloc() and/or realloc(), expand the dynamic array as needed, and ....

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Thanks. I have been reading about these functions:

1) When using malloc like in

Code:

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#include <stdio.h>
#include <stdlib.h>

int main()
{

int *temp;

temp = malloc(sizeof(int));
temp[0]=1;
temp[1]=2;
temp[2]=3;

return 0;
}

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an error is not produced. Why is that? The command malloc(sizeof(int)) only reserves memory for 1 integer, but my array can continue beyond that?

2) Is there any logic behind the fact that I do not need to typecast the void-pointer malloc returns? Or is this just how C works?

3) When I use the free()-command, I free the piece of memory malloc has reserved. Does this mean that the void-pointer orignally returned by malloc cannot point to the memory anymore?

Best,
Niles.

Quote:

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Originally Posted by Niels_M

an error is not produced. Why is that? The command malloc(sizeof(int)) only reserves memory for 1 integer, but my array can continue beyond that?

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You're accessing the array out of bounds.

Quote:

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Originally Posted by Niels_M

2) Is there any logic behind the fact that I do not need to typecast the void-pointer malloc returns? Or is this just how C works?

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It is just how C works.

Quote:

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Originally Posted by Niels_M

3) When I use the free()-command, I free the piece of memory malloc has reserved. Does this mean that the void-pointer orignally returned by malloc cannot point to the memory anymore?

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It still contains the address, except that the address is no longer that of the dynamic array because the dynamic array is no more. Maybe it will get reused for another dynamic array, but maybe not.

Quote:

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Originally Posted by laserlight

1)You're accessing the array out of bounds.


2)It is just how C works.


3)It still contains the address, except that the address is no longer that of the dynamic array because the dynamic array is no more. Maybe it will get reused for another dynamic array, but maybe not.

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Thanks. Regarding realloc, from http://www.cplusplus.com/reference/c...dlib/realloc/:

"The function may move the memory block to a new location, in which case the new location is returned."

If that is the case (i.e. that the memory block may be moved, then I should free both "data" and "temp" when I have the following piece of code, right?

Code:

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temp=realloc(data, 2*sizeof(char));

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Quote:

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Originally Posted by Niels_M

If that is the case (i.e. that the memory block may be moved, then I should free both "data" and "temp" when I have the following piece of code, right?

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No, if necessary, realloc() will deallocate the old dynamic array for you.

Please RTFM.

Quote:

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Originally Posted by laserlight

No, if necessary, realloc() will deallocate the old dynamic array for you.

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Thanks.

Quote:

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Originally Posted by Bayint Naung

Please RTFM.

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I will, sorry.

Beware that if realloc fails, it would return NULL. And that means if you blindly assign the return of realloc to your pointer that you got from malloc, you will overwrite the old value and get a memory leak. Be careful.
If realloc fails, possibly free what you got from malloc.

Quote:

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Originally Posted by Elysia

Beware that if realloc fails, it would return NULL. And that means if you blindly assign the return of realloc to your pointer that you got from malloc, you will overwrite the old value and get a memory leak. Be careful.
If realloc fails, possible free what you got from malloc.

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Note that that cplusplus.com article's example makes this mistake, although it is forgivable by virtue of the fact that the program just bails out by ending its life (seppuku?) should realloc return a null pointer.