Trying to understand pointers for function paramaters

[Richard Hengeveld]

Hi all,

I'm trying to understand how pointers for function parameters work. As I
understand it, if you got a function like:

void f(int *i)
{
*i = 0;
}

int main()
{
int a;
f(&a);
return 0;
}

It does what you want (namely altering the value of a).
I find this illogical. As far as I can understand, the address of "a" is
passed, and "*i" is set with this address not "i", as it should be in my
understanding.
What am I missing?
TIA

P.S.
I've really searched for this in the groups faq and elsewhere before I
posted.

 

[Mark A. Odell]

Hi all,

I'm trying to understand how pointers for function parameters work. As I
understand it, if you got a function like:

void f(int *i)
{
*i = 0;
}

int main()
{
int a;
f(&a);
return 0;
}

Assume that 'a' exist at memory location 0x1000'0000. Now assume 'i' is
located at 0x1000'0004.

Before calling f():
-------------------
C name Mem. Addr. Contents (value located there)
------ ------------ ------------------------------
a 0x1000'0000: ???
i 0x1000'0004: ???

In function f() after the assignment to *i:
-------------------------------------------

C name Mem. Addr. Contents (value located there)
------ ------------ ------------------------------
a 0x1000'0000: 0
i 0x1000'0004: 0x1000'0000

so 'i' is a pointer that contains the value of the address of 'a'. Thus
weh you dereference 'i' via *i you now point to memory location
0x1000'0000. So if you modify what is as 0x1000'0000 then the value of 'a'
will be modified.

Simple, logical.

 

[Keith Thompson]

Hi all,

I'm trying to understand how pointers for function parameters work. As I
understand it, if you got a function like:

void f(int *i)
{
*i = 0;
}

int main()
{
int a;
f(&a);
return 0;
}

It does what you want (namely altering the value of a).
I find this illogical. As far as I can understand, the address of "a" is
passed, and "*i" is set with this address not "i", as it should be in my
understanding.
What am I missing?

Yes, you're passing the address of "a" to the function "f".

Incidentally, "i" is a poor name for the parameter. "i" is commonly
used as a name for an int variable; the parameter is a pointer to int.
It's perfectly legal, but potentially confusing.

In the assignment

*i = 0;

"i" is a pointer, and "*i" is an int object. You're assigning the
value 0 to an int object. Which int object? The one i points to,
which happens to be "a".

If you had written

i = 0;

you'd be assigning a value to "i", which is a pointer object; the
value being assigned would be a null pointer.

It might be clearer if you change the names:

void f(int *ptr_param)
{
*ptr_param = 0;
}

int main(void)
{
int int_object;
f(&int_object);
return 0;
}

(In a real program, of course, variables should generally have names
that reflect what they're used for; for this toy example, it's more
important to show their types.)

 

[E. Robert Tisdale]

I'm trying to understand how pointers for function parameters work.
As I understand it, if you got a function like:

void f(int* p) {
*p = 0;
}

int main(int argc, char* argv[]) {
int a;
f(&a);
return 0;
}

It does what you want (namely altering the value of a).
I find this illogical. As far as I can understand,
the address of "a" is passed,
and "*p" is set with this address not "p"
as it should be in my understanding.

Probably what is confusing you is the *formal* argument

int* p

This tells you that p is a pointer to an object of type int.
the statement

p = 0;

would assign the address value 0 to p.
The expression

*p

is a *reference* to (another name for) a
so writing

*p = 0;

is the same thing as writing

a = 0;

 

[Richard Hengeveld]

I'm trying to understand how pointers for function parameters work.
As I understand it, if you got a function like:

void f(int* p) {
*p = 0;
}

int main(int argc, char* argv[]) {
int a;
f(&a);
return 0;
}

It does what you want (namely altering the value of a).
I find this illogical. As far as I can understand,
the address of "a" is passed,
and "*p" is set with this address not "p"
as it should be in my understanding.

Probably what is confusing you is the *formal* argument

int* p

Thanks for replying.
Yes, that is exactly what is confusing me.
I understand you set a pointer (if you're not passing to functions) by:

int a, *p;
p = &a;

and not:

p* = &a

Wouldn't it be more logical if it was something like:

void f(int i) {
int *p
p = i;
p* = 0;
}

int main(int argc, char* argv[]) {
int a;
f(&a);
return 0;
}

?

 

[Keith Thompson]

I'm trying to understand how pointers for function parameters work.
As I understand it, if you got a function like:

void f(int* p) {
*p = 0;
}

int main(int argc, char* argv[]) {
int a;
f(&a);
return 0;
}

It does what you want (namely altering the value of a).
I find this illogical. As far as I can understand,
the address of "a" is passed,
and "*p" is set with this address not "p"
as it should be in my understanding.

Damn it, Tisdale, that's not what he wrote. Here's what Richard
Hengeveld *actually* wrote in the article to which you replied:

] I'm trying to understand how pointers for function parameters work. As I
] understand it, if you got a function like:
]
] void f(int *i)
] {
] *i = 0;
] }
]
] int main()
] {
] int a;
] f(&a);
] return 0;
] }
]
] It does what you want (namely altering the value of a).
] I find this illogical. As far as I can understand, the address of "a" is
] passed, and "*i" is set with this address not "i", as it should be in my
] understanding.

By prefixing the material with "> ", you're telling us that you're
quoting what the previous poster wrote; by preceding it with "Richard
Hengeveld wrote:", you're saying so explicitly. In fact, you're
showing us your version of what you think he should have written.
You've quietly changed the layout of his code (I like his better),
changed the name of the parameter from "i" to "p", and added
gratuitous argc and argv parameters to main.

Now if you had presented your modified version as an improvement of
Richard Hengeveld's code (which is what I did in my response), that
would have been ok. If you had mentioned that you were paraphrasing
what he posted rather than quoting it exactly, that would have been
acceptable as well. You could even have told us why you think your
modified version is an improvement. (In some ways it is, but that's
beside the point.)

You've done this before, and you've been called on it. I have no
realistic expectation that you're going to change your ways this time
either. I'm posting this mostly as a warning to other readers.

If E. Robert Tisdale claims that someone else has written something in
a previous article, don't believe it unless you've verified it by
reading the actual article that he claims to be quoting. Trust him at
your own risk.

 

[Default User]

Richard Hengeveld wrote:

Wouldn't it be more logical if it was something like:

void f(int i) {
int *p
p = i;
p* = 0;

How would it be more logical to assign an integer to a
pointer-to-integer? What problem are you trying to solve?




Brian Rodenborn

 

[Richard Hengeveld]

Richard Hengeveld wrote:




How would it be more logical to assign an integer to a
pointer-to-integer? What problem are you trying to solve?

I'm not trying to solve a problem. I'm just trying to understand C.
In a function without pointer arguments, the argument(s) of that function
are set by value passing:

f(int i)
{
printf("%d", i);
}

int main()
{
int i;
f(i);
return 0;
}

I just don't understand why this is (a little bit) different with pointer
arguments.

 

[E. Robert Tisdale]

I understand you set a pointer (if you're not passing to functions) by:

int a, *p;
p = &a;

and not:

p* = &a

Wouldn't it be more logical if it was something like:

void f(int i) {
int *p
p = i;
p* = 0;
}

int main(int argc, char* argv[]) {
int a;
f(&a);
return 0;
}

?

I don't know.
I don't know why K&R chose these semantics
and not the semantics that you suggest.
suppose that you wanted to define a second pointer
to the same object. Would you write

int *p, *q;
p = i;
q = i;

And would you expect (p == q) to be true?

 

[Keith Thompson]

Thanks for replying.
Yes, that is exactly what is confusing me.
I understand you set a pointer (if you're not passing to functions) by:

int a, *p;
p = &a;

Right, this sets p to contain the address of a (or, equivalently,
causes p to point to a).

and not:

p* = &a

The unary '*' operator is prefix, not postfix. You could argue that
it would be easier if it were postfix, but that's not how the language
defines it.

If you meant
*p = &a;
that wouldn't make sense. Since p is a pointer to int, *p is an int,
but &a (address of a) is a pointer to int. The types don't match.

In an assignment, the left hand side has to be an expression (e.g., a
name) that refers to an object of some type, and the right hand side
has to be an expression of that same type (not necessarily an object).
(That's not quite true (there are implicit conversions in some cases),
but it's a good enough first approximation.)

So given:
int a;
int *p;
we can have
a = 42; /* the name "a" refers to an object of type int,
"42" is an expression of type int */
p = &a; /* the name "p" refers to an object of type pointer-to-int,
"&a" is an expression of type pointer-to-int */
*p = 2+2; /* the name "*p" refers to an object of type int
(the object happens to be "a"), and "2+2" is an expression
of type int */

Wouldn't it be more logical if it was something like:

void f(int i) {
int *p
p = i;

No, p refers to an object of type pointer-to-int, but i is an expression
of type int. The types don't match. You can legally say "p = &i;".

p* = 0;

p* is a syntax error. *p = 0; is legal, since *p refers to an object
of type int (assuming p has been initialized properly), and 0 is an
expression of type int.

}

int main(int argc, char* argv[]) {
int a;
f(&a);

The function f() expects an argument of type int; you're giving it an
argument of type pointer-to-int. Function argument types have to
match, just as the types in an assignment have to match.

return 0;
}

If you want the function f() to be able to modify an int object that
you pass to it, you have to pass the object's address, and f() has to
take an argument of type pointer-to-int. If f() takes an argument of
type int, it just gets a copy of the value of whatever you pass to it;
f() can do whatever it likes with its own copy, but that won't affect
the original object.

If f() is defined as:

void f(int i) { ... whatever ... }

then this:

int x;
f(x);

can't change the value of x, any more than this:

f(42);

can change the value of 42.

Some languages do have ways of specifying that an argument is passed
in a way that allows the function to modify the original object
(Pascal has VAR parameters, Ada as "in out" and "out" parameters, C++
has reference parameters). C doesn't have such a mechanism.

If you want to argue that C could be improved, there are several
things I can say in response:

1. You're right. C's declaration syntax in particular causes no end
of headaches, especially to novices.

2. It's not going to change. Any significant change would break
existing code. That's just not going to happen. If you want a
language with better syntax than C (and that's a subjective
judgement), you'll just have to use a language other than C.

3. If you're interested in C, you should probably learn the language
as it is before you start worrying about how it could be better. K&R2
(Kernighan & Ritchie's _The C Programming Language_, Second Edition)
is one of the best tutorials; buy or borrow a copy and read it.

 

[Default User]

I'm not trying to solve a problem. I'm just trying to understand C.
In a function without pointer arguments, the argument(s) of that
function are set by value passing:

f(int i)
{
printf("%d", i);
}

int main()
{
int i;
f(i);
return 0;
}

I just don't understand why this is (a little bit) different with
pointer arguments.

It's not different. What makes you think it is? In both cases,
something is passed by value (that means a copy of it was made) and
assigned to a local variable, the parameter.

It just so happens that in one case it was a integer, and the other it
was a pointer-to-integer. In the second case, a copy of the address was
made and passed to the function. It's still the same address that was
created through the use of the & operator in main().

You don't appear to understand how pointers work. This is fundamental
to the use of the C language. Get a good book and read it. Otherwise,
you are wasting your time and ours.




Brian Rodenborn

 

[pete]

As far as I can understand, the address of "a" is
passed, and "*i" is set with this address not "i",
as it should be in my understanding.

After
int *i = &a
you have
i == &a
not
*i == &a

 

[Pedro Graca]

Thanks for replying.
Yes, that is exactly what is confusing me.

[newbie answer -- maybe this helps]


int i, *p; // declare an int and a pointer to int
void f(int *p); // declare a function taking a pointer to int parameter

I think you'd understand that function better if you'd write it as

void g(int* p); // make 'int*' stand out



typedef int *pointer_to_int;
typedef int* pointer_to_int;

These two typedefs are absolutely equal. Which one do you prefer?

void h(pointer_to_int p);

 

[CBFalconer]

.... snip ...

Incidentally, "i" is a poor name for the parameter. "i" is
commonly used as a name for an int variable; the parameter is a
pointer to int. It's perfectly legal, but potentially confusing.

In the assignment

*i = 0;

"i" is a pointer, and "*i" is an int object. You're assigning the
value 0 to an int object. Which int object? The one i points to,
which happens to be "a".

If you had written

i = 0;

you'd be assigning a value to "i", which is a pointer object; the
value being assigned would be a null pointer.

It might be clearer if you change the names:

void f(int *ptr_param)
{
*ptr_param = 0;
}

int main(void)
{
int int_object;
f(&int_object);
return 0;
}

(In a real program, of course, variables should generally have names
that reflect what they're used for; for this toy example, it's more
important to show their types.)

While I abhor Hungarian notation, I do like to append either p or
ptr to the names of pointer variables.

 

[CBFalconer]

.... snip mangled quote ...

Damn it, Tisdale, that's not what he wrote. Here's what Richard
Hengeveld *actually* wrote in the article to which you replied:
.... snip details ...

You've done this before, and you've been called on it. I have no
realistic expectation that you're going to change your ways this
time either. I'm posting this mostly as a warning to other readers.

If E. Robert Tisdale claims that someone else has written something
in a previous article, don't believe it unless you've verified it
by reading the actual article that he claims to be quoting. Trust
him at your own risk.

And just when I thought Trollsdale was showing signs of
reformation. He seems to be a dedicated recidivist. Can we apply
the 'three strikes' law somehow?

 

[Barry Schwarz]

Thanks for replying.
Yes, that is exactly what is confusing me.
I understand you set a pointer (if you're not passing to functions) by:

int a, *p;
p = &a;

and not:

p* = &a

Wouldn't it be more logical if it was something like:

void f(int i) {
int *p
p = i;

I assume you meant p=&i here.

p* = 0;

I assume you meant *p=0 here. This will set i to 0 but I is local to
f. When you return, i ceases to exist. Consequently, the argument in
the calling function is never updated and defeats the purpose.

}

int main(int argc, char* argv[]) {
int a;
f(&a);

With f as defined above, this obviously causes a diagnostic since f is
expecting an int but you are passing an int*.

Going back to your original post which defined f as expecting an int*,
when you call f with this argument, three significant things happen:

i is created as an automatic variable local to f.
i is initialized with the address of a (which is a local variable
local to main.
Control is transferred to f.

Look back at your code at the top of this post and notice how similar
the first two are to what you wrote when not calling a function.

return 0;
}

The concepts to understand are

In a definition, *i defines i as a pointer.

After the definition, i refers to the pointer itself and not the
object being pointed to, if any.

Once the pointer has been initialized, coding *i dereferences the
pointer and evaluates to the object being pointed to.

A function call causes all the parameters of the function to be
initialized with the corresponding arguments used in the calling
statement. (So f(&a) has a very similar effect to p=&a.)


<<Remove the del for email>>

 

[Flash Gordon]

And just when I thought Trollsdale was showing signs of
reformation. He seems to be a dedicated recidivist. Can we apply
the 'three strikes' law somehow?

You mean if any of us meet him we get to strike him three times?

 

[Mark A. Odell]

I'm not trying to solve a problem. I'm just trying to understand C.
In a function without pointer arguments, the argument(s) of that
function are set by value passing:

Correct. In this case the called function gets a "copy" of the value to
play with and the caller can be sure that the callee won't modify the
original value in 'i'. This is nice when you want to preserve your value
but have some function create a new value based on your value, e.g.

#include <stdio.h>

static int plusTwo(int copyOfVal)
{
/* Modify copy by adding two to demonstrate
** that the caller's 'val' is not modified.
*/
copyOfVal += 2;

return copyOfVal;
}

static int timesTenModify(int *pVal)
{
int failure = 0;

if (pVal) *pVal *= 10;
else failure = !0;

return failure;
}

int main(void)
{
int failure;
int val = 10;
int valPlusTwo = plusTwo(val);

/* As the caller, I retain my original value of 'val'
** whilst having a worker function calculate 'val' + 2
** for me which I store in 'valPlusTwo'. In this situation
** I do not wish my 'val' to be modified.
*/
printf("My val is %d, and +2 is %d\n", val, valPlusTwo);

/* In this case, I wish to modify 'val' and do not need
** to retain its original value of 10 so I call
** timesTenModify() with a pointer to my 'val' object
** so it can modify 'val' directly.
*/
failure = timesTenModify(&val);
if (failure) printf("Trouble with timesTenModify() call\n");
else printf("My val is now %d\n", val);

return 0;
}

I just don't understand why this is (a little bit) different with
pointer arguments.

Hopefully this simple example will help.

 

[Christopher Benson-Manica]

And just when I thought Trollsdale was showing signs of
reformation. He seems to be a dedicated recidivist. Can we apply
the 'three strikes' law somehow?

Perhaps there could be a section of the FAQ dedicated to identifying
posters to plonk?

 

[Christopher Benson-Manica]

void g(int* p); // make 'int*' stand out

That's actually misleading - it makes "int*" appear to be a type,
which it is not.

typedef int *pointer_to_int;
typedef int* pointer_to_int;

These two typedefs are absolutely equal. Which one do you prefer?

Neither, for real code.