const parameter problem - clarification needed

[subramanian100in]

Consider the following program:

#include <stdio.h>

void myfn(const int **a)
{
static int i, j, k;

a[0] = &i;
a[1] = &j;
a[2] = &k;

return;
}

int main(void)
{
int *a[3];
int x, y, z;

a[0] = &x;
a[1] = &y;
a[2] = &z;

x = 10;
y = 20;
z = 30;

myfn(a);

return 0;

}

Suppose the program name is tmp.c

When this program is compiled with gcc under Redhat Enerprise Linux,
with the command
gcc -std=c99 tmp.c

the following Compilation warning is produced with gcc
const_ptr.c: In function `main':
const_ptr.c:27: warning: passing arg 1 of `myfn' from incompatible
pointer type

If I remove the const qualifier in myfn( ), the program compiles with
gcc without any warning.

However with VC++ 2005, there is no compilation warning with the
original program ie even when the const qualifier is present.

QUESTION:
-----------------
Why is the warning generated with gcc ?

This question arises because of the following reason.
We can pass a char[] to a function which receives it as const char*.
Here there is no compilation warning reported. But when a char *a[] is
passed, why can't it be received as "const char ** " ?

[fdmfdmfdm]

On Feb 19, 7:11 am, "subramanian10. [e-mail], India"
<subramanian10. [e-mail] te:

Consider the following program:

#include <stdio.h>

void myfn(const int **a)
{
static int i, j, k;

a[0] = &i;
a[1] = &j;
a[2] = &k;

return;

}

int main(void)
{
int *a[3];
int x, y, z;

a[0] = &x;
a[1] = &y;
a[2] = &z;

x = 10;
y = 20;
z = 30;

myfn(a);

return 0;

}

Suppose the program name is tmp.c

When this program is compiled with gcc under Redhat Enerprise Linux,
with the command
gcc -std=c99 tmp.c

the following Compilation warning is produced with gcc
const_ptr.c: In function `main':
const_ptr.c:27: warning: passing arg 1 of `myfn' from incompatible
pointer type

If I remove the const qualifier in myfn( ), the program compiles with
gcc without any warning.

However with VC++ 2005, there is no compilation warning with the
original program ie even when the const qualifier is present.

QUESTION:
-----------------
Why is the warning generated with gcc ?

This question arises because of the following reason.
We can pass a char[] to a function which receives it as const char*.
Here there is no compilation warning reported. But when a char *a[] is
passed, why can't it be received as "const char ** " ?

I think the problem is if you are using "const int **", you are
telling the compiler that you are passing a pointer of pointer
pointing to a constant integer value. But in your main, "a" is not
pointing to a const value. This will have problem.

[Andrey Tarasevich]

su************* *@yahoo.com, India wrote:

...
void myfn(const int **a)
{
...
}

int main(void)
{
int *a[3];
...
myfn(a);
...
}

...
the following Compilation warning is produced with gcc
const_ptr.c: In function `main':
const_ptr.c:27: warning: passing arg 1 of `myfn' from incompatible
pointer type

If I remove the const qualifier in myfn( ), the program compiles with
gcc without any warning.
...

Argument 'a' in the function call has type 'int*[3]' which in this particular
context decays to type 'int**'. The corresponding parameter has type 'const
int**', which means that in this case you are trying to convert a 'int**' value
to 'const int**' type implicitly. This is not legal in C language. 'int*' can be
implicitly converted to 'const int*', but 'int**' cannot be implicitly converted
to 'const int**'. It the latter conversion were legal, it would let one to
circumvent const-correctness rules without an explicit cast. See the following
FAQ item for more detail

http://www.parashift.com/c++-faq-lit...html#faq-18.17

(This is a C++ FAQ entry, but the principle it illustrates applies immediately
to C as well)

--
Best regards,
Andrey Tarasevich