arrays - where they finish?

[Olaf \El Blanco\]

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

 

[Vladimir S. Oka]

Olaf "El Blanco" opined:

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

You can't, in general case.

If your arrays are terminated with a specific value, like strings are
with '\0' in C, you could traverse them until reaching that value.

If they were declared like:

int array[SOME_SIZE];

you could get their size like this:

sizeof array / sizeof array[0] ( == SOME_SIZE)

There is no other way in C.

 

[Olaf \El Blanco\]

Olaf "El Blanco" opined:

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

You can't, in general case.

If your arrays are terminated with a specific value, like strings are
with '\0' in C, you could traverse them until reaching that value.

If they were declared like:

int array[SOME_SIZE];

No, is declared like:
int *array

So, If I want to print all the array, I can't???

 

[santosh]

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

Your question is not clear. If you want to know the number of elements
in a fixed size array then sizeof(array)/sizeof(array[0]) should yield
the answer. If it a dynamically allocated array, then you'll have to
keep track of it's size, (and hence the number of elements it has)
yourself. For example if *array is of type pointer to long and it's
initialised to point to the start of an array of N bytes then array
points to N/sizeof *array elements that you can access.

Hope this helps somewhat.

 

[santosh]

Olaf "El Blanco" opined:

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

You can't, in general case.

If your arrays are terminated with a specific value, like strings are
with '\0' in C, you could traverse them until reaching that value.

If they were declared like:

int array[SOME_SIZE];

No, is declared like:
int *array

So, If I want to print all the array, I can't???

'array' as shown above is merely a variable of type pointer to int. It
will not point to any legally accessible location unless you initialise
it somehow, either via malloc() or assigning the result of the &
operator on some object. How big that object or region of memory is, is
completely upto you to know and keep track of. C has no provisions for
automatically doing it for you. If you don't know that a pointer won't
automatically point to a legal range of memory, you should pick up a
good book on C and start at the beginning.

 

[Kenneth Brody]

Olaf "El Blanco" opined:

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)

[...]
You can't, in general case.

If your arrays are terminated with a specific value, like strings are
with '\0' in C, you could traverse them until reaching that value.

If they were declared like:

int array[SOME_SIZE];

No, is declared like:
int *array

So, If I want to print all the array, I can't???

Not unless something else has stored/passed you the length, or you have
agreed that some "magic value" is stored in the last entry. (For example,
in an array of "char*"s, you might add a NULL on the end.)

--
+-------------------------+--------------------+-----------------------------+
| Kenneth J. Brody | www.hvcomputer.com | |
| kenbrody/at\spamcop.net | www.fptech.com | #include <std_disclaimer.h> |
+-------------------------+--------------------+-----------------------------+
Don't e-mail me at: <mailto:[email protected]>

 

[Keith Thompson]

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)

"int *array" declare int as a pointer, not as an array; similarly for
"char *array". If you have a pointer to the first element of an array
(which is a very common idiom), there's no direct way to compute the
size of the array it points to.

The comp.lang.c FAQ is at <http://www.c-faq.com/>. Read section 6,
"Arrays and Pointers". If you still have questions after that, feel
free to come back and ask them.

 

[John F]

Olaf "El Blanco" opined:

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

You can't, in general case.

If your arrays are terminated with a specific value, like strings
are
with '\0' in C, you could traverse them until reaching that value.

If they were declared like:

int array[SOME_SIZE];

No, is declared like:
int *array

So, If I want to print all the array, I can't???

You will need to "remember" the number of elements... like

int array_el_cnt=...;

regards
John

 

[John Bode]

I want to know how many integers has 'array'. (int *array)
Or chars has 'array' (char *array)
Etc...


So...
while (array) { ... } ????
while (array != EOF) {...} ????
while (array != NULL) {...} ???

(I want to use the minimun possible variables)

If you have the array definition in scope:

int a[10];
size_t asize;
...
/*
* Divide the total number of bytes used by a
* by the number of bytes used in a single
* element of a
*/
asize = sizeof a / sizeof a[0];

asize contains the number of elements (10) in the array.

Note that this *will not* work if the array is passed as a function
parameter; because of how C treats arrays, what actually gets passed is
a pointer to the first element of the function. If you're passing an
array to a function, you must also pass the number of elements in the
array as a separate parameter.

One workaround for this is to wrap an array in a struct type and pass
the struct as the parameter; it's not as convenient as it sounds,
however.

If you've declared the array dynamically through malloc() or calloc(),
you have to keep track of the number of elements you allocated.

 

[Olaf \El Blanco\]

THANKS and sorry!
Is just I saw some really small codes and I want to "copy" the style...
But I understand... My questions was ridiculous!
int *array
Of course we don't know how many integers are in that part of the memory...
just go up and check which number appear in malloc!

Thanks again.

 

[santosh]

THANKS and sorry!

.... snip ...

Quite okay, but do take time to read the following:

<http://cfaj.freeshell.org/google/>
<http://clc-wiki.net/wiki/Introduction_to_comp.lang.c>
<http://www.safalra.com/special/googlegroupsreply/>

 

[August Karlstrom]

THANKS and sorry!
Is just I saw some really small codes and I want to "copy" the style...
But I understand... My questions was ridiculous!

Not at all. Many languages keep track of the length of arrays, even
dynamically allocated ones.

int *array
Of course we don't know how many integers are in that part of the memory...
just go up and check which number appear in malloc!

Thanks again.

August