Array and Pointer Tutorial

T

Tomás

Some programmers treat arrays just like pointers (and some even think that
they're exactly equivalent). I'm going to demonstrate the differences.

Firstly, let's assume that we're working on a platform which has the
following properties:

1) char's are 8-Bit. ( "char" is synomonous with "byte" ).
2) int's are 32-Bit. ( sizeof(int) == 4 ).
3) Pointers are 64-Bit. ( sizeof(int*) == 8 ).


First let's make two declarations:

int main(void)
{
int array[5];

int* const pointer = (int*)malloc( 5 * sizeof(int) );
}


Now I'll demonstrate how "array" and "pointer" are different:


I'll start off with simple analgous expressions:

============================================================================
| Expression | Type and Access Specifiers | That in English |
============================================================================
| | | |
| array | int[5] | An array of five int's.|
| | | |
|---------------------------------------------------------------------------
| | |A const pointer which |
| pointer | int* const |points to a modifiable |
| | |int. |
|--------------------------------------------------------------------------|
| | |A const pointer which |
| &array | int (* const)[5] |points to a modifiable |
| | |array of five int's. |
|--------------------------------------------------------------------------|
| | |A const pointer, which |
| &pointer | int* const* const |points to a const |
| | |pointer, which points to|
| | |a modifiable int. |
============================================================================


Here's how "sizeof" works with them:


===========================================================
| Expression | sizeof( exp ) | But Why? |
===========================================================
| | | |
| array | 20 | It's five int's. |
| | (5 * 4) | |
|---------------------------------------------------------|
| | | |
| pointer | 8 | It's just a pointer.|
| | (just 8) | |
|---------------------------------------------------------|
| | | |
| &array | 8 | It's just a pointer.|
| | (just 8) | |
|----------------------------------------------------------
| | | |
| &pointer | 8 | It's just a pointer.|
| | | |
| | (just 8) | |
===========================================================


Okay next thing to discuss is the usage of square brackets, and the
dereference operator. The two of these are to be used by pointers only. So
how come we can use them with arrays, as follows?:

array[0] = 4;

*array = 6;

The reason is that an expression of the following type:

int[5]

can implicitly convert to an expression of the following type:

int* const

What it does is convert to a pointer to the first element of the array.
Therefore, the first example:

array[0] = 4;

implicitly converts "array" to a normal pointer, then uses chain brackets to
access memory at a certain offset from the original address.

Also the second example:

*array = 6;

implicitly converts "array" to a normal pointer, then dereferences it.


NOTE: You must remember that an array implicitly converts to a pointer to
its first element, NOT to a pointer to the array. This fact has a few
implications. Here's one such implication:


*(array + 3) = 6;


What the above line of code does is the following:

1) Implicitly converts "array" to an: int* const
2) Adds 3 * sizeof(int) to the address.
3) Dereferences the resulting pointer, and assigns 6 to it.


If "array" implicitly converted to: int (*)[5]
rather than a pointer to the first element, then Step 2 above would be
different, specifically:


2) Adds 3 * sizeof( int[5] ) to the address.


And we know that sizeof(int[5]) is 20 on this platform (not 8!).


So you may ask, "What's the point in having a pointer to an array?" -- well
here's where it may come in handy:


void SomeFunc ( int (* const p_array)[5] )
{
(*p_array)[0] = 99;
(*p_array)[1] = 98;
(*p_array)[2] = 97;
(*p_array)[3] = 96;
(*p_array)[4] = 95;

/* This function won't accept an array of any
other size! */
}


And here's a C++-specific example with references:

void SomeFunc ( int (&array)[5] )
{
array[0] = 99;
array[1] = 98;
array[2] = 97;
array[3] = 96;
array[4] = 95;

/* This function won't accept an array of any
other size! */
}


Also in C++, you can exploit the use of templates:

template<class T, unsigned long len>
void KeepCopy( const T (&array)[len] )
{
static my_array[len];

/* Do some other stuff */
}


I've posted this to a few newsgroups, so if you'd like to reply, please post
to comp.lang.c because it's the common denominator. If your post is C++-
specific, the please post to comp.lang.c++.

Did I leave anything out?

-Tomás

[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]
 
R

Richard Bos

Tomás said:
Some programmers treat arrays just like pointers (and some even think that
they're exactly equivalent). I'm going to demonstrate the differences.

Firstly, let's assume that we're working on a platform which has the
following properties:

1) char's are 8-Bit. ( "char" is synomonous with "byte" ).

A char is _always_ a byte. What you mean is that you are assuming a char
(and therefore also a byte) to be equal to an octet.
2) int's are 32-Bit. ( sizeof(int) == 4 ).
3) Pointers are 64-Bit. ( sizeof(int*) == 8 ).

First let's make two declarations:

int main(void)
{
int array[5];

int* const pointer = (int*)malloc( 5 * sizeof(int) );

*Boing* and here the demonstration crashes.

Never cast malloc(). It is not necessary. void *s can be freely
converted to and from any object pointer type.
Never use malloc() (or any other function, for that matter) without a
proper declaration in scope. It forces your compiler to assume that the
function returns an int, which malloc() clearly does not.

Note that the first error hides the second. The combination of the two
can result in garbage being assigned to your pointer.

As a matter of convenient maintenance, not a true error, it is more
dabble-proof to use sizeof *pointer rather than sizeof (type). If your
pointer's type changes, the first form stays correct, the second can
turn deceptively (and hiddenly!) broken.

All in all, that program should've looked like this:

#include <stdlib.h>

int main(void)
{
int array[5];

int* const pointer = malloc(5 * sizeof *pointer);
}

Anyway, the difference between pointers and arrays is most simply
demonstrated using the age[1]-old method of arrows and groups of boxes.

Richard

[1] I.e., in the computing world, a couple of decades
 
?

=?ISO-8859-15?Q?Juli=E1n?= Albo

Tomás said:
Some programmers treat arrays just like pointers (and some even think that
they're exactly equivalent). I'm going to demonstrate the differences.

Some of these days you can also write an article demonstrating that integer
types are not floating point types. It will be another great contribution
to the progress of humanity.
 
C

Chad

As a matter of convenient maintenance, not a true error, it is more
dabble-proof to use sizeof *pointer rather than sizeof (type). If your
pointer's type changes, the first form stays correct, the second can
turn deceptively (and hiddenly!) broken.

All in all, that program should've looked like this:

#include <stdlib.h>

int main(void)
{
int array[5];

int* const pointer = malloc(5 * sizeof *pointer);
}

Anyway, the difference between pointers and arrays is most simply
demonstrated using the age[1]-old method of arrows and groups of boxes.

Richard

[1] I.e., in the computing world, a couple of decades

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

return 0;
}


Now, I change
int array[5];

to

double array[5];

Wouldn't the sizeof double be truncated to the sizeof int? If so, the
wouldn't this create an additional bug?

Chad
 
C

CBFalconer

Chad said:
.... snip ...

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

There is no such thing as *array. array is an array of 5 integers,
not a pointer.

--
Some informative links:
http://www.geocities.com/nnqweb/
http://www.catb.org/~esr/faqs/smart-questions.html
http://www.caliburn.nl/topposting.html
http://www.netmeister.org/news/learn2quote.html
 
F

Flash Gordon

Chad said:
As a matter of convenient maintenance, not a true error, it is more
dabble-proof to use sizeof *pointer rather than sizeof (type). If your
pointer's type changes, the first form stays correct, the second can
turn deceptively (and hiddenly!) broken.

All in all, that program should've looked like this:

#include <stdlib.h>

int main(void)
{
int array[5];

int* const pointer = malloc(5 * sizeof *pointer);
}

Anyway, the difference between pointers and arrays is most simply
demonstrated using the age[1]-old method of arrows and groups of boxes.

Richard

[1] I.e., in the computing world, a couple of decades

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

return 0;
}


Now, I change
int array[5];

to

double array[5];

Wouldn't the sizeof double be truncated to the sizeof int? If so, the
wouldn't this create an additional bug?

You have not followed the suggested practice of using sizeof *pointer
being assigned to. Your example should have been:
#include <stdlib.h>

int main(void) {
int array[5];

int *q = malloc(sizeof *q);

return 0;
}

Or, if you want array and q to always be the same type, you could have used:
#include <stdlib.h>
int main(void) {
int array[5], *q = malloc(sizeof *q);

return 0;
}

I really can't see why you would think of using sizeof some other object.
--
Flash Gordon, living in interesting times.
Web site - http://home.flash-gordon.me.uk/
comp.lang.c posting guidelines and intro:
http://clc-wiki.net/wiki/Intro_to_clc

Inviato da X-Privat.Org - Registrazione gratuita http://www.x-privat.org/join.php
 
C

Charles Richmond

CBFalconer said:
... snip ...

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

There is no such thing as *array. array is an array of 5 integers,
not a pointer.
"array[0]" is the same as "*(array+0)" which simplifies to "*array".
 
C

CBFalconer

Charles said:
CBFalconer said:
Chad said:
... snip ...

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

There is no such thing as *array. array is an array of 5 integers,
not a pointer.

"array[0]" is the same as "*(array+0)" which simplifies to "*array".

When array is passed as a parameter, which this one isn't.

--
Some informative links:
http://www.geocities.com/nnqweb/
http://www.catb.org/~esr/faqs/smart-questions.html
http://www.caliburn.nl/topposting.html
http://www.netmeister.org/news/learn2quote.html
 
P

pete

CBFalconer said:
Chad wrote:
int array[5];

int *q = malloc(sizeof *array);

There is no such thing as *array. array is an array of 5 integers,
not a pointer.

(*array) is an expression of type int.

The operand of sizeof is (*array), not (array),
so, array gets converted to a pointer.
 
T

Tomás

CBFalconer posted:

"array[0]" is the same as "*(array+0)" which simplifies to "*array".

When array is passed as a parameter, which this one isn't.


Stop trolling and propogating misinformation.

"array" implicitly converts to a pointer to its first element ALL THE TIME
-- NOT just when passed as an argument to a function, NOT just when it's a
global variable, NOT just when you have porridge instead of cereal.

-Tomás
 
N

Nelu

Tomás said:
CBFalconer posted:

"array[0]" is the same as "*(array+0)" which simplifies to "*array".

When array is passed as a parameter, which this one isn't.


Stop trolling and propogating misinformation.

"array" implicitly converts to a pointer to its first element ALL THE TIME
-- NOT just when passed as an argument to a function, NOT just when it's a
global variable, NOT just when you have porridge instead of cereal.

I keep trying to compile the following program:

void f(char *b) {
b++;
}

int main(void) {
char a[]={'a','b','c'};

a++;

f(a);

return 0;
}

Let me know where it fails on your C compiler as a should implicitly
convert to a pointer, right?
 
P

pete

Nelu said:
Tomás said:
CBFalconer posted:

"array[0]" is the same as "*(array+0)"
which simplifies to "*array".

When array is passed as a parameter, which this one isn't.


Stop trolling and propogating misinformation.

"array" implicitly converts to a pointer to its first element
ALL THE TIME
-- NOT just when passed as an argument to a function,
NOT just when it's a
global variable, NOT just when you have porridge instead of cereal.

Stop trolling and propogating misinformation.

An expression of array type is not coverted
when it is the operand of the sizeof operator.
An expression of array type is not coverted
when it is the operand of the address operator.
I keep trying to compile the following program:
int main(void) {
char a[]={'a','b','c'};

a++;
Let me know where it fails on your C compiler as a should implicitly
convert to a pointer, right?

"(8) : error C2105: '++' needs l-value"

One problem with (a++),
is that the result of a type conversion is not an lvalue.

N869
6.3.2 Other operands
6.3.2.1 Lvalues and function designators

[#3] Except when it is the operand of the sizeof operator or
the unary & operator, or is a string literal used to
initialize an array, an expression that has type ``array of
type'' is converted to an expression with type ``pointer to
type'' that points to the initial element of the array
object and is not an lvalue. If the array object has
register storage class, the behavior is undefined.
 
N

Nelu

pete said:
Nelu said:
Tomás said:
CBFalconer posted:


"array[0]" is the same as "*(array+0)"
which simplifies to "*array".

When array is passed as a parameter, which this one isn't.


Stop trolling and propogating misinformation.

"array" implicitly converts to a pointer to its first element
ALL THE TIME
-- NOT just when passed as an argument to a function,
NOT just when it's a
global variable, NOT just when you have porridge instead of cereal.

Stop trolling and propogating misinformation.

An expression of array type is not coverted
when it is the operand of the sizeof operator.
An expression of array type is not coverted
when it is the operand of the address operator.
I keep trying to compile the following program:
int main(void) {
char a[]={'a','b','c'};

a++;
Let me know where it fails on your C compiler as a should implicitly
convert to a pointer, right?

"(8) : error C2105: '++' needs l-value"

One problem with (a++),
is that the result of a type conversion is not an lvalue.

N869
6.3.2 Other operands
6.3.2.1 Lvalues and function designators

[#3] Except when it is the operand of the sizeof operator or
the unary & operator, or is a string literal used to
initialize an array, an expression that has type ``array of
type'' is converted to an expression with type ``pointer to
type'' that points to the initial element of the array
object and is not an lvalue. If the array object has
register storage class, the behavior is undefined.

If it's not an lvalue doesn't it mean that it's not a pointer *object*
but a pointer value?
 
T

Tomás

If it's not an lvalue doesn't it mean that it's not a pointer *object*
but a pointer value?


From the Standard:

An lvalue refers to an object or function. Some rvalue expressions —-
those of class or cv-qualified class type -- also refer to objects.


-Tomás
 
P

pete

Nelu said:
Nelu said:
Tomás wrote:
CBFalconer posted:


"array[0]" is the same as "*(array+0)"
which simplifies to "*array".

When array is passed as a parameter, which this one isn't.


Stop trolling and propogating misinformation.

"array" implicitly converts to a pointer to its first element
ALL THE TIME
-- NOT just when passed as an argument to a function,
NOT just when it's a
global variable, NOT just when you have porridge instead of cereal.

Stop trolling and propogating misinformation.

An expression of array type is not coverted
when it is the operand of the sizeof operator.
An expression of array type is not coverted
when it is the operand of the address operator.
I keep trying to compile the following program:
int main(void) {
char a[]={'a','b','c'};

a++;
Let me know where it fails on your C compiler
as a should implicitly
convert to a pointer, right?

"(8) : error C2105: '++' needs l-value"

One problem with (a++),
is that the result of a type conversion is not an lvalue.

N869
6.3.2 Other operands
6.3.2.1 Lvalues and function designators

[#3] Except when it is the operand of the sizeof operator or
the unary & operator, or is a string literal used to
initialize an array, an expression that has type ``array of
type'' is converted to an expression with type ``pointer to
type'' that points to the initial element of the array
object and is not an lvalue. If the array object has
register storage class, the behavior is undefined.

If it's not an lvalue doesn't it mean that it's not a pointer *object*
but a pointer value?

Yes.

(a++) means the same thing as ((a + 0)++)
 
P

pete

Tomás said:
Some rvalue expressions —-
those of class or cv-qualified class type -- also refer to objects.

What does "class or cv-qualified class type" mean?

Is that C++?
 
C

Chad

You have not followed the suggested practice of using sizeof *pointer
being assigned to. Your example should have been:
#include <stdlib.h>

int main(void) {
int array[5];

int *q = malloc(sizeof *q);

return 0;
}

Or, if you want array and q to always be the same type, you could have used:
#include <stdlib.h>
int main(void) {
int array[5], *q = malloc(sizeof *q);

return 0;
}

I really can't see why you would think of using sizeof some other object.
--
Flash Gordon, living in interesting times.
Web site - http://home.flash-gordon.me.uk/
comp.lang.c posting guidelines and intro:
http://clc-wiki.net/wiki/Intro_to_clc

Inviato da X-Privat.Org - Registrazione gratuita http://www.x-privat.org/join.php

I can't really put me pinpoint the exact part, but I know I'm missing
some kind of underlying concept when I see the construction:

int *q = malloc(sizeof *q);

Maybe when I'm scrubbing the dairy coolers at work tomorrow morning,
the exact part that is irking me will set in. Then I can come back and
ask the EXACT question about the construction.

Chad
 
R

Richard Heathfield

Chad said:
I can't really put me pinpoint the exact part, but I know I'm missing
some kind of underlying concept when I see the construction:

int *q = malloc(sizeof *q);

The canonical way to allocate space for n objects of type T is:

T *p = malloc(n * sizeof *p);

or, if p is already declared, simply this:

p = malloc(n * sizeof *p);

The reason this is the canonical way is that it doesn't rely on the type of
p, except that it must be an object type, not an incomplete type or
function type. If the type of p changes during maintenance, you don't have
to hunt down this line and hack it about. It will automagically work
correctly with the new type.
 
J

Jack Klein

Chad said:
... snip ...

Okay, I'm probably missing this. But say I have the following:

/*I omitted checking for NULL and using free*/

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

int main(void) {
int array[5];

int *q = malloc(sizeof *array);

There is no such thing as *array. array is an array of 5 integers,
not a pointer.

I beg your pardon. There most definitely is such a thing as *array,
because of the implicit and automatic conversion of the name of an
array to a pointer in all expressions other than unary & and sizeof.

Compile and execute yourself:

#include <stdio.h>

int main(void)
{
double array1 [10];
char array2 [10];

printf("sizeof *array1 = %d\n", (int)sizeof *array1);
printf("sizeof *array2 = %d\n", (int)sizeof *array2);

return 0;
}

I get:

sizeof *array1 = 8
sizeof *array2 = 1
 
C

Chad

Richard said:
Chad said:

The canonical way to allocate space for n objects of type T is:

T *p = malloc(n * sizeof *p);

or, if p is already declared, simply this:

p = malloc(n * sizeof *p);

The reason this is the canonical way is that it doesn't rely on the type of
p, except that it must be an object type, not an incomplete type or
function type. If the type of p changes during maintenance, you don't have
to hunt down this line and hack it about. It will automagically work
correctly with the new type.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Hmmm... It looks like got a response before I could correct the grammar
errors in my previous post. Anyhow, I think I managed to pinpoint what
is irking me.

Say I have the following:

/*I omitted checking malloc() for NULL and forgot to use free()*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
char *p = malloc(sizeof *p);
/* incomplete -- malloc's return value not checked */
strcpy(p, "Hello, world!");

return 0;
}

Are you saying that the construction
char *p = malloc(sizeof *p);

would allocate enough space to hold the string ""Hello, world!" versus
allocating memory to hold just one char?

Chad
 

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