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char** function parameters

Suppose I have the following function:

void doSomething(char** array, int size);

Normally, I would call the function with an array of, say, 3
elements...

char* strs[] = { "str1","str2","str3" };
doSomething(strs, 3);

That works fine. But if I want to pass it only one char*, I would
think this would work....

char* str = "str1";
doSomething(&str, 1);

It compiles with a warning about incompatible pointer types, and
needless to say it doesn't behave as I expected. "str1" does not get
passed into the function. What am I doing wrong? How can I pass a
single char* into the function?

Nov 15 '05 #1
12 2621
On Tue, 19 Jul 2005 co***********@hotmail.com wrote:
Suppose I have the following function:

void doSomething(char** array, int size);

Normally, I would call the function with an array of, say, 3
elements...

char* strs[] = { "str1","str2","str3" };
doSomething(strs, 3);

That works fine. But if I want to pass it only one char*, I would
think this would work....

char* str = "str1";
doSomething(&str, 1);

It compiles with a warning about incompatible pointer types, and
needless to say it doesn't behave as I expected. "str1" does not get
passed into the function. What am I doing wrong? How can I pass a
single char* into the function?


Here is a really simple program. There isn't a compilation problem :

/* Compiled with gcc -Wall -pedantic -O3 */
#include <stdio.h>
#include <stdlib.h>

void f(char **t)
{
printf("%s\n", *t);
}

int main(void)
{
char * str = "str0";
char * tab[] = { "str1", "str2", "str3" };
f(&str);
f(tab);
return EXIT_SUCCESS;
}

Everything goes fine.

So my guess is you're having another kind of problem, which isn't directly
related to passing your pointer.

--
"Je deteste les ordinateurs : ils font toujours ce que je dis, jamais ce
que je veux !"
"The obvious mathematical breakthrough would be development of an easy
way to factor large prime numbers." (Bill Gates, The Road Ahead)
Nov 15 '05 #2
co***********@hotmail.com wrote:
Suppose I have the following function:

void doSomething(char** array, int size);

Normally, I would call the function with an array of, say, 3
elements...

char* strs[] = { "str1","str2","str3" };
doSomething(strs, 3);

That works fine. But if I want to pass it only one char*, I would
think this would work....

char* str = "str1";
doSomething(&str, 1);

It compiles with a warning about incompatible pointer types, and
needless to say it doesn't behave as I expected. "str1" does not get
passed into the function. What am I doing wrong? How can I pass a
single char* into the function?


I don't see anything wrong with what you posted, can you post a small
but complete program that produces the warning when compiled, and what
exactly is not behaving as you expected?

Robert Gamble

Nov 15 '05 #3
The problem was that I was treating a char[256] as a char*. Here is a
modification of Stephane's program to demonstrate what I want to do:

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

void f(char **t)
{
printf("%s\n", *t);
}

int main(void)
{
char foo[256];
strcpy(foo, "bar");

f(&foo); /* WRONG */

return EXIT_SUCCESS;
}

The gcc compiler warning is "assignment from incompatible pointer
type," and what prints out when I run the program is garbage.

So is it possible to pass the contents of foo[256] into the function
f()?

Nov 15 '05 #4
co***********@hotmail.com wrote:
The problem was that I was treating a char[256] as a char*. Here is a
modification of Stephane's program to demonstrate what I want to do:

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

void f(char **t)
{
printf("%s\n", *t);
}

int main(void)
{
char foo[256];
strcpy(foo, "bar");
you need to #include <string.h> for strcpy.
f(&foo); /* WRONG */
Yes, that is wrong. f takes an argument of type pointer to pointer to
char, you are passing pointer to char (foo and &foo decay into a
pointer to the first element of the array, it is the same as &foo[0]).
return EXIT_SUCCESS;
}

The gcc compiler warning is "assignment from incompatible pointer
type," and what prints out when I run the program is garbage.

So is it possible to pass the contents of foo[256] into the function
f()?


You can do this:

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

void f(char **t)
{
printf("%s\n", *t);
}
int main(void)
{
char foo[256];
char *ptr = foo;
strcpy(foo, "bar");
f(&ptr)
return EXIT_SUCCESS;
}

Robert Gamble

Nov 15 '05 #5
Robert Gamble wrote:

f(&foo); /* WRONG */


Yes, that is wrong. f takes an argument of type pointer to pointer to
char, you are passing pointer to char (foo and &foo decay into a
pointer to the first element of the array, it is the same as &foo[0]).

That's incorrect. &foo is a pointer to the array, so it has type
pointer to array 256 of char, NOT pointer to char.


Brian
Nov 15 '05 #6


Robert Gamble wrote:
co***********@hotmail.com wrote:
The problem was that I was treating a char[256] as a char*. Here is a
modification of Stephane's program to demonstrate what I want to do:

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

void f(char **t)
{
printf("%s\n", *t);
}

int main(void)
{
char foo[256];
strcpy(foo, "bar");

you need to #include <string.h> for strcpy.

f(&foo); /* WRONG */

Yes, that is wrong. f takes an argument of type pointer to pointer to
char, you are passing pointer to char (foo and &foo decay into a
pointer to the first element of the array, it is the same as &foo[0]).


No, no, no. Please see Question 6.12 in the comp.lang.c
Frequently Asked Questions (FAQ) list

http://www.eskimo.com/~scs/C-faq/top.html

In particular,

- Plain `foo' decays to a pointer to the array's first
element, but `&foo' does not.

- Plain `foo' is the same as `&foo[0]', but `&foo' is
a different thing altogether.

- A proper understanding of the type of `&foo' is the
key to the O.P.'s question.

--
Er*********@sun.com

Nov 15 '05 #7


co***********@hotmail.com wrote:
The problem was that I was treating a char[256] as a char*. Here is a
modification of Stephane's program to demonstrate what I want to do:

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

void f(char **t)
{
printf("%s\n", *t);
}

int main(void)
{
char foo[256];
strcpy(foo, "bar");

f(&foo); /* WRONG */

Ah, enlightenment dawns. Try something like this:

char *p = foo;
f(&p);
return EXIT_SUCCESS;
}

The gcc compiler warning is "assignment from incompatible pointer
type," and what prints out when I run the program is garbage.

Yes, because the type of &foo is char(*)[256], not char**.
So is it possible to pass the contents of foo[256] into the function
f()?


Try the trick above, see if it helps.

Nov 15 '05 #8
Default User wrote:
Robert Gamble wrote:

f(&foo); /* WRONG */


Yes, that is wrong. f takes an argument of type pointer to pointer to
char, you are passing pointer to char (foo and &foo decay into a
pointer to the first element of the array, it is the same as &foo[0]).

That's incorrect. &foo is a pointer to the array, so it has type
pointer to array 256 of char, NOT pointer to char.


Yup, thanks for the correction.

Robert Gamble

Nov 15 '05 #9
>co***********@hotmail.com wrote:
[edited for space]
void f(char **t) { printf("%s\n", *t); }
int main(void) {
char foo[256];
strcpy(foo, "bar");
f(&foo); /* WRONG */

In article <11**********************@g49g2000cwa.googlegroups .com>
John Bode <jo*******@my-deja.com> wrote:Ah, enlightenment dawns. Try something like this:

char *p = foo;
f(&p);


Indeed. A picture might also help:

foo:
+-----+-----+-----+-----+-----...-----+
| 'b' | 'a' | 'r' | 0 | (junk)... |
+-----+-----+-----+-----+-----...-----+

Here "foo" is an array of size 256 containing "char"s. Note that
there *is no pointer*, there is just the array named "foo", which
occupies 256 bytes.

Inside f(), assuming that **t == 'k' and that the printf()
will print (say) "k2":

t:
+----------------------+ +----------------------+
| *------------------> | * |
+----------------------+ +-----------|----------+
/
/
/
|
v
+-----+-----+-----+
| 'k' | '2' | 0 |
+-----+-----+-----+

Here there are *two* pointers: t, and *t. In this illustration
I drew them both as four bytes long (by making the boxes about
four times the size of the one-byte "char" boxes), but they could
be 2 or 3 or 4 or 8 or 128 bytes, or (on some rather unusual
systems) even just one byte. The important item is that there
are, and *must be*, two pointers -- the one named t, and one that
t points to -- before you can use **t. (The call to printf()
will access (*t)[0], aka **t, and then (*t)[1], and then (*t)[2],
and so on.)

Now, if you just use "foo" in a value context (or write &foo[0]),
the compiler will *construct* a pointer value, pointing to the
first element of the array:

foo:
+-----+-----+-----+-----+-...-+
*------> | 'b' | 'a' | 'r' | 0 | ... |
+-----+-----+-----+-----+-...-+

but this pointer is not (necessarily) stored in memory anywhere,
as it is a mere value, not an object. By adding "char *p = foo",
we create an actual object, so now the above becomes:

p: foo:
+----------+ +-----+-----+-----+-----+-...-+
| *------> | 'b' | 'a' | 'r' | 0 | ... |
+----------+ +-----+-----+-----+-----+-...-+

Now if we call f(&p), we pass to f() a value pointing to the pointer
named "p". f()'s first actions, even before any code inside f()
gets excuted, are to copy that value into an object, the one we
named "t" in f(). In other words, function parameters are really
just ordinary local variables, initialized "by magic" as we begin
executing the function, using the values passed in from the caller.
So now we have "t" (in f()) pointing to "p" (in the main() that I
snipped) pointing to &foo[0] (also in main()), so now the picture
we *wanted* -- t pointing to a pointer that points to the first of
a series of "char"s -- is in fact the picture we *have*.

Whenever you (the generic "you") are struggling with pointers, it
can help to draw pictures. Each named object (variable) is a box
containing a value, or junk if it is uninitialized. If the type
of the object is "pointer to ...", the object contains an arrow.
You need to make sure the arrow points in turn to some other thing
somewhere in memory -- another named object, or perhaps memory
obtained from malloc(). Each chunk of memory in turn contains a
value (or junk), and if the type it is meant to contain is "pointer
to ...", the value is an arrow -- and you have to make the arrow
point somewhere useful, just like last time. You can use a pointer
to write to the thing to which the pointer points, so one
arrow pointing to the first of several uninitialized-memory arrows
can be used to initialize them:

T **ptr = malloc(2 * sizeof *ptr);

produces (assuming malloc() succeeds) this picture:

(pointing off into the weeds)
ptr: /
+---------+ +----/---+---------+
| *---------------> | * | * |
+---------+ +--------+-----\---+
\ (more weeds)

and now you can say "ptr[0] = (some expression)" to set ptr[0],
and likewise with ptr[1]. As before, you can point them to
named variables (of type T), or call malloc() again. The
important thing is to make them point somewhere, and be sure
they still point somewhere valid whenever you use them.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (4039.22'N, 11150.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
Nov 15 '05 #10
Primarily what you need to do is to avoid the warning you need to
typecast
by doing doSomething((char **)&str,1);
So let's say your function is

void doSomething(char** s, int size)
{
printf("%s\n",*s);
}
This needs to be done because you are not printing s[0], instead are
trying to print address of s[0];
Remember this is a double pointer.

Nov 15 '05 #11
> Primarily what you need to do is to avoid the warning you need to
typecast
by doing doSomething((char **)&str,1);


I'm not sure I'm understanding why you would have to absolutely try to
hush the compiler. Warnings are there for a good reason, most of the time.
The only thing we need to know is "is this warning announcing something
that was unexpected ?"... well, IMO at least :)

--
"Je deteste les ordinateurs : ils font toujours ce que je dis, jamais ce
que je veux !"
"The obvious mathematical breakthrough would be development of an easy
way to factor large prime numbers." (Bill Gates, The Road Ahead)
Nov 15 '05 #12
Thanks for all the feedback!

Nov 15 '05 #13

This discussion thread is closed

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