Using const qualifier

[amvoiepd]

Hi,
My question is about how to use const properly.
I have two examples describing my problem.

First, let's say I have a linked list and from it I want to
find some special node. I write the function, and then figure
that the function will not be modifying the list at all, so a
const qualifier seems appropriate in the parameter.
So essentially I have:

struct node {
...
struct node *next;
};

struct node *find_special(const struct node *n)
{
while (...) {
...
n = n->next;
....
}
return n;
}

But this will give me a warning since I am returning n,
which is declared const, and my function is not returning
a const. How do I fix this? If I change my function to
return a const, I would be stuck with a variable which I can
not change when I use my function, right?

Also, let's say I have a function that assigns a pointer to a
member of a struct. Here I also think a const would do:

void assign_pointer(struct mystruct *s, const int *p)
{
s->pointer = p;
}

struct mystruct {
int *pointer;
};

Same problem here, I'm assigning a const to a non const.
What to do? I can't add const to my struct member, since
I want to do things with it later.

So how do I use const properly? I do understand the reason
why these examples fail, just not how to handle it.

Thanks!

 

[Peter Nilsson]

Hi,
My question is about how to use const properly.
I have two examples describing my problem.

First, let's say I have a linked list and from it I want to
find some special node. I write the function, and then figure
that the function will not be modifying the list at all, so a
const qualifier seems appropriate in the parameter.
So essentially I have:

struct node {
    ...
    struct node *next;
};

struct node *find_special(const struct node *n)
{
    while (...) {
        ...
        n = n->next;
        ....
    }
    return n;
}

But this will give me a warning since I am returning n,
which is declared const, and my function is not returning
a const. How do I fix this?

Cast the return value to non-const.

If I change my function to
return a const, I would be stuck with a variable which I can
not change when I use my function, right?
Yes.

Also, let's say I have a function that assigns a pointer to a
member of a struct. Here I also think a const would do:

void assign_pointer(struct mystruct *s, const int *p)

This would only be a valid design if 's->pointer' is also const.
If it isn't (and isn't meant to be) remove the const
qualifier against *p.

 

[amvoiepd]

Cast the return value to non-const.

I knew that was one possible way to silence the warning, but
having read this groups for some time, I had the impression that
casts are almost never necessary. Is this a case where a cast
is perfectly okay?

This would only be a valid design if 's->pointer' is alsoconst.
If it isn't (and isn't meant to be) remove theconstqualifieragainst *p.

What makes the first example a valid design, but not this one if 's-

pointer'

is not const? I mean, in respective function the pointers are not
altered
at all, but after the function ends they can be altered. I don't see
the
difference between them. Does a function need to take into account
that a
pointer may be changed after returning from a function?

Thanks!

 

[Eric Sosman]

[... function deriving non-const value from const argument ...]
But this will give me a warning since I am returning n,
which is declaredconst, and my function is not returning
aconst. How do I fix this?

Cast the return value to non-const.

I knew that was one possible way to silence the warning, but
having read this groups for some time, I had the impression that
casts are almost never necessary. Is this a case where a cast
is perfectly okay?

Yes, I'd say so. C's system of types is not very
flexible, and the type qualifiers like const can become
inconvenient at times, providing useful documentation for
a parameter at the cost of "poisoning" other expressions.
Cast it away.

As for "casts are almost never necessary," I think
that's a misstatement of "most casts are misused." The
latter isn't a statement about casts, really, but about
the programmers who misuse them -- and whose numbers are
legion. In English, "comprise" is almost always wrong
not because there's anything amiss with the word, but
because it is nearly always misused. If you can use
"comprise" correctly you are in a minority; a similar
situation seems to hold with casts.

 

[gw7rib]

Hi,
My question is about how to use const properly.
I have two examples describing my problem.

What Peter and Eric have said elsethread is correct, but I think your
conception may be wrong.

First, let's say I have a linked list and from it I want to
find some special node. I write the function, and then figure
that the function will not be modifying the list at all, so a
const qualifier seems appropriate in the parameter.
So essentially I have:

struct node {
    ...
    struct node *next;

};

struct node *find_special(const struct node *n)
{
    while (...) {
        ...
        n = n->next;
        ....
    }
    return n;

}

This function returns a non-const pointer to a node of the list. The
function itself does not modify the list. But it returns a pointer
through which you could modify the list. So your function, having
promised not to modify the list, can't then guarantee that the promise
is kept. This is why you are having problems.

Also, let's say I have a function that assigns a pointer to a
member of a struct. Here I also think a const would do:

void assign_pointer(struct mystruct *s, const int *p)
{
    s->pointer = p;

}

struct mystruct {
    int *pointer;

};

Similar problem here. Your function promises that it won't alter the
int that p points to. But it lets "pointer" point to it, and "pointer"
is free to alter the value pointed at.

I think you need to ask yourself the question "Do I want the pointed-
at thing to be alterable, or not?" and arrange your consts
accordingly.

Hope that helps.
Paul.

 

[Chris Torek]

[The node-finding] function returns a non-const pointer to a
node of the list. The function itself does not modify the list.
But it returns a pointer through which you could modify the list.
So your function, having promised not to modify the list, can't
then guarantee that the promise is kept. This is why you are
having problems.

Indeed: given a function that does not modify the object or objects
to which its argument pointer(s) point, but *does* return a pointer
to one of those objects, you have a problem.

Do you use "const" to imply "I will not change this, and thus you
can pass to me a pointer to an actually-const, really-in-ROM object",
and then have the return value "de-const-ified"? In which case,
you can do bad things:

typedef <whatever> T;
T *deconstify(const T *);
const T read_only_obj = { value_that_cannot_change };
*deconstify(&read_only_obj) = oops_a_bug;

Or, do you avoid const, after which you cannot even *call* the
function even though you would have done it safely? For instance:

T *nonconst(T *); /* nonconst() does not modify it though */
const T *ptr;
ptr = nonconst(&read_only_obj); /* alas, this gets a diagnostic */

The Standard C Library takes the former approach: functions like
strchr() "deconstify" their argument:

const char ro[] = "room!";

*strchr(ro, 'r') = 'b'; /* boom! */

(which may indeed "go boom", i.e., crash, and does on some of my
systems).

One of the many reasons C++ has overloaded functions is to sidestep
this problem. In C++, the types of arguments -- including whether
they are const-qualified -- determine which function is actually
called, so we simply write two completely separate functions:

const int *operate(const int *);
int *operate(int *);

Then:

const int obj1 = 0
int obj2 = 0;

*operate(&obj1) = 42;

will not compile, because &obj1 has type "const int *", so this
calls operate(const int *), which returns "const int *", not
"int *". Of course, this opens the door to abuse: the two
separate functions might well do entirely different things.
Ideally, the const-qualified version of the function does the
same thing as the non-const-qualified version, and in a language
better than C++, we could avoid writing two functions by simply
declaring that the return type matches the argument type:

flexible operate(arg) [
constraint: typeof(operate) is typeof(arg);
constraint: typeof(arg) is choice { int *, const int * };
] {
...
return arg;
}

which also of course lets us write "overloaded" functions without
repeating them six ways from Sunday:

/*
* Substring operation: find "needle" in "haystack"
*/
flexible substring(haystack, needle) [
constraint: typeof(operate) is typeof(haystack);
constraint: typeof(needle) is typeof(haystack);
constraint: typeof(haystack) is choice {
char *, const char *,
Unicode16 *, const Unicode16 *,
Unicode32 *, const Unicode32 *
};
] {
... code to find substring ...
}

(To achieve the above, C++ generally uses "templates" instead of,
or combined with, "overloaded" functions, but there is really
no need for both.)

(I made up the above syntax on the fly, so it is probably full
of flaws.)

 

[Army1987]

Chris Torek wrote:

[snip]

Ideally, the const-qualified version of the function does the
same thing as the non-const-qualified version, and in a language
better than C++, we could avoid writing two functions by simply
declaring that the return type matches the argument type: [snip]
which also of course lets us write "overloaded" functions without
repeating them six ways from Sunday:

/*
* Substring operation: find "needle" in "haystack"
*/
flexible substring(haystack, needle) [
constraint: typeof(operate) is typeof(haystack);
constraint: typeof(needle) is typeof(haystack);

Why? What's wrong with allowing substring(L"unicode string", "string")?

constraint: typeof(haystack) is choice {
char *, const char *,
Unicode16 *, const Unicode16 *,
Unicode32 *, const Unicode32 *
};
] {
... code to find substring ...
}