char** function parameters

C

confusedcoder

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?
 
S

Stephane Zuckerman

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.
 
R

Robert Gamble

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
 
C

confusedcoder

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()?
 
R

Robert Gamble

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 said:
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
 
D

Default User

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
 
E

Eric Sosman

Robert said:
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 */


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.
 
J

John Bode

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.
 
R

Robert Gamble

Default said:
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
 
C

Chris Torek

(e-mail address removed) wrote:
[edited for space]
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);

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.
 
R

Rajan

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.
 
S

Stephane Zuckerman

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 :)
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

No members online now.

Forum statistics

Threads
473,744
Messages
2,569,484
Members
44,904
Latest member
HealthyVisionsCBDPrice

Latest Threads

Top