incompatibility of argument with prototype

[przemek drochomirecki]

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

thx in adv.

Przemek Drochomirecki

[Rolf Magnus]

przemek drochomirecki wrote:

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

The types don't match. foo takes a pointer to pointer to const char, but you
pass a pointer to pointer to non-const char.

[przemek drochomirecki]

Rolf Magnus wrote:

przemek drochomirecki wrote:

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

The types don't match. foo takes a pointer to pointer to const char, but you
pass a pointer to pointer to non-const char.

I know that types don't match, but...still don't know why...:)

int foo(char* const* x) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

Here everything is OK...
Thx, anyway.

Przemek Drochomirecki

[Victor Bazarov]

przemek drochomirecki wrote:

[...]

I know that types don't match, but...still don't know why...:)

It's actually explained in the FAQ. See section 18.

V

[Maxim Yegorushkin]

On Sat, 18 Jun 2005 01:43:04 +0400, przemek drochomirecki
<dr*****@nospam.hyene.com> wrote:

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

thx in adv.

Przemek Drochomirecki

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

--
Maxim Yegorushkin

[Greg]

Rolf Magnus wrote:

przemek drochomirecki wrote:

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

The types don't match. foo takes a pointer to pointer to const char, but you
pass a pointer to pointer to non-const char.

....which of course will compile without a problem. It's always
acceptable to pass non-const types when const types are declared. It's
the opposite situation, passing const types as non-const parameters
that the compiler will complain about. But clearly that is not a
problem in this code.

In fact, there is no error in the sample code - a fact which could well
explain the original poster's difficulty in understanding the error.

Greg

[Alf P. Steinbach]

* Greg:

Rolf Magnus wrote:

przemek drochomirecki wrote:

Hi,

I don't understand the idea behind the error in the following code...
int foo(const char **p) { return 0; }

int main(int argc, char **argv)
{
foo(argv);
return 0;
}

The types don't match. foo takes a pointer to pointer to const char, but you
pass a pointer to pointer to non-const char.

In fact, there is no error in the sample code - a fact which could well
explain the original poster's difficulty in understanding the error.

In fact the code is invalid.

The reason it is invalid is that doing what the code attempts to allows
treating a const object as a non-const one, without a cast.

As Victor wrote elsewhere in this thread, see the FAQ.

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