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How does one code something like this?

 
 
Daniel Rudy
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      01-02-2006

Hello,

I have a peice of code that I'm making an attempt to code. The problem
is that I need to return an arbitrary number of char strings.

int function(char *fname, int *dcount, char *data[])

Would this work? If so, then how to you load the data into data? Malloc?


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Daniel Rudy

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WaJo
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      01-02-2006
Hi,

I guess data should be declared like this Char ** data,
yes, malloc will do it...

Regards,

 
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Emmanuel Delahaye
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      01-02-2006
Daniel Rudy a écrit :
> I have a peice of code that I'm making an attempt to code. The problem
> is that I need to return an arbitrary number of char strings.
>
> int function(char *fname, int *dcount, char *data[])
>
> Would this work?


Nope.

char **function(char const *fname, int *dcount)

or better

struct data
{
size_t count;
char **arr;
};

int function (char const *fname, struct data *p_data)

--
A+

Emmanuel Delahaye
 
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Malcolm
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      01-02-2006

"Daniel Rudy" <(E-Mail Removed)> wrote
> I have a peice of code that I'm making an attempt to code. The problem
> is that I need to return an arbitrary number of char strings.
>
> int function(char *fname, int *dcount, char *data[])
>
> Would this work? If so, then how to you load the data into data? Malloc?
>

You can't do it that way.
The reason is that you don't know how many strings you have. Therefore you
need to call

char **ptr;
int N; / *set to number of strings */
ptr = malloc(N * sizeof(char *));

(in fact you would probably malloc() 1 and then call realloc() as you add
strings to the list, but that's just a detail).

The pointer therefore changes, so you can't pass it it. You need a char ***,
if you want to return it as a parameter.
A triple pointer is getting messy, so it's probably better to return the
pointer

char **function(char *fname, int *dcount)


 
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Daniel Rudy
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      01-02-2006
At about the time of 1/2/2006 2:26 PM, Malcolm stated the following:
> "Daniel Rudy" <(E-Mail Removed)> wrote
>
>>I have a peice of code that I'm making an attempt to code. The problem
>>is that I need to return an arbitrary number of char strings.
>>
>>int function(char *fname, int *dcount, char *data[])
>>
>>Would this work? If so, then how to you load the data into data? Malloc?
>>

>
> You can't do it that way.
> The reason is that you don't know how many strings you have. Therefore you
> need to call
>
> char **ptr;
> int N; / *set to number of strings */
> ptr = malloc(N * sizeof(char *));
>
> (in fact you would probably malloc() 1 and then call realloc() as you add
> strings to the list, but that's just a detail).
>
> The pointer therefore changes, so you can't pass it it. You need a char ***,
> if you want to return it as a parameter.
> A triple pointer is getting messy, so it's probably better to return the
> pointer
>
> char **function(char *fname, int *dcount)
>
>


I'm having trouble visualizing this. I thought that char **ptr is a
pointer to a pointer to a char (or array of chars). Am I wrong? I know
that it can be done, look at int main(int argc, char *argv[]) which
everyone knows. Hmm... I'm going to abandon this method and try
something else, which may also result in another post on declaring
nested unions and structures.

--
Daniel Rudy

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Malcolm
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      01-03-2006

"Daniel Rudy" <(E-Mail Removed)> wrote
> I'm having trouble visualizing this. I thought that char **ptr is a
> pointer to a pointer to a char (or array of chars). Am I wrong? I know
> that it can be done, look at int main(int argc, char *argv[]) which
> everyone knows. Hmm... I'm going to abandon this method and try
> something else, which may also result in another post on declaring
> nested unions and structures.
>

A pointer can point to anything.

If it points to an integer it is an int *, if it points to a character it is
a char *.
It can also point to another pointer. So a pointer to a pointer that points
to a character would be a char **. A pointer to that would be a char ***,
and so on (in practise you harldy ever need more than three levels).

In C, you work with arrays of values by pointing to the first element. So if
we have three integer in memory

12 15 -4

We can set a pointer, call it ptr, to point to the 12.

*ptr gives us the 12
*(ptr +1) gives us the 15
*(ptr +2) gives us the -4.

We can also use array notation

ptr[0] is exactly the same a *ptr
ptr[1] is exactly the same as (*ptr+1)

This is especially useful for strings, because we generally don't know how
long they will be. So you simply point to the first character in memory

a s t r i n g 0

we point to the 'a', and then we can examine all the other characters. The 0
tells us where the string ends.




 
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Keith Thompson
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      01-03-2006
"Malcolm" <(E-Mail Removed)> writes:
[...]
> We can also use array notation
>
> ptr[0] is exactly the same a *ptr
> ptr[1] is exactly the same as (*ptr+1)


Quibble: ptr[1] is the same as *(ptr+1).

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Markus Becker
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      01-03-2006
Hi, I think you're mostly right, but you wrote it in a way, that
could easily be mis-interpreted, especially by someone who
needs to have this explained ...

Malcolm <(E-Mail Removed)> schrieb:

> A pointer can point to anything.


Better: Pointers can point to any type of object in memory.
You can't point it to, say, the controller of your hard disk


> If it points to an integer it is an int *, if it points to a character it is
> a char *.


To further clarify, this does *not* mean that an int pointer
that accidentally points to a char object suddenly changes
its type from int* to char*. An int* is an int* and will remain
an int* as long as the program is running.

An expample:

struct { char str[10]; int i[10];} array;
char *pc = &array.str[9]; // takes the address of the last element in str
int *pi = &array.i; // address of the first (zero'th) element in i

Now incrementing pc like in:

pc = pc + 1;

does not change the type of the pointer. pc is still a char*
and when you try to do something with it, you get undefined
behaviour. The operation to increment a pointer to one place
_behind_ the last element of an array is perfectly legal and
well defined.

On the other hand:

pi = pi - 1;

does not change the type of pi from int* to char*, in fact
(even if it were allowed and defined what to do when you de-
crement a pointer to a place before the beginning of an array,
which it isn't) does not even point to str[9] but probably
to str[7]. In real life it can point anywhere, because of
padding bytes the compiler puts between the end of str[] and
the beginning of i[] to ensure that the elements contained
in i[] can be accessed (some systems cannot access integers
that reside on odd adresses, some can but extremely slow,
some will cause bus errors or protection faults, ....
it's undefined, _anything_ can happen).

> It can also point to another pointer.


Additionally, a void* can point to _any_ type of object.
But that's about all that a void* can, or what you can
do with a void*. a void* can point to a void** (can it?)
but I have yet to discover of what use a void** could be.

Before you use it for something more trivial than just
pointing around, e.g. dereferencing it to access the object
it points to, you have to cast it to a pointer of a
particular type, meaning that _you_ have to know the type
of object it points to. The compiler cannot do that, it
relies on your knowing what you are doing. On some systems
(especially embedded controllers) this can be used for
some nasty tricks, but in 'standard' C this leads to
undefined behaviour.

The advantage of void* is that they can be
casted to any other type and the accesses made via the casted
(and therefore typed) pointer are guaranteed to succed in re-
retrieving a porperly typed object at the location it points
to. If you point a void* to the address of str[1] in the above
struct of my 'example' and then cast it to an int*, you'll
get undefined behaviour on most modern systems where
sizeof(char) != sizeof(int). The result of this misaligned
pointing is undefined, it you do nothing more than point
to objects which, on a particular system, cannot be stored
at this location nothing serious will happen, but if you
try to access the misaligned object, your program will
most certainly some kind of access error that will crash
your program.

to cast a void* to some other type of pointer, in C you
do not need the explicit cast like

int i;
void* pv=&i;
int *pi=(int*)pv;

it is sufficient to just write:

int *pi = pv;

The compiler selflessly does the conversion itself ('implicit'
cast).

> *ptr gives us the 12
> *(ptr +1) gives us the 15
> *(ptr +2) gives us the -4.


Note that we increment the pointer by the number of objects.

The compiler knows their size and appropriately ensures that
the increment in memory locations is correct.
This is the reason why one can't do pointer-arithmetic
with void* pointers, because the objects the pointer points
to have no known type.
You can add pointers and integers, but not pointers and
pointers. With one exception: if both pointers are of the
same type (e.g. int*) and both point to locations inside
an int-array (up to one position after the last element
of the array) you can use p2-p1 to get the number of
elements between p1 and p2.

> We can also use array notation
>
> ptr[0] is exactly the same a *ptr
> ptr[1] is exactly the same as (*ptr+1)


Typo: ptr[1] == *(ptr+1)

> This is especially useful for strings, because we generally don't know how
> long they will be.


Which is one of the reasons why one shouldn't use functions like
gets() or strcpy().
Oops, different thread.

> a s t r i n g 0


better: "astring\0"

> we point to the 'a', and then we can examine all the other characters. The 0
> tells us where the string ends.


Markus, hoping that I didn't make too much mistakes at this
early time of day...
 
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