«

Hi All,
I have one question regarding the code.


#include<stdio.h>
char *f1(void);
char *f1(void)
{
char *abc ="Hello";

return abc;
}

int main(void )
{
char *p;
p = f1();

printf("%s\n", p);
return 0;
}


While compiling the above program as mentioned below I am getting the
meaning
$ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
jj.c
jj.c: In function `f1':
jj.c:5: warning: initialization discards qualifiers from pointer
target type

I have two question
1) What is the meaning of the warning ?
2) is it safe to return the local pointer value (i.e return abc is
correct ?

My understanding is we should not return address of local
variable .So above code may not be working always .

Regards,
Somenath

somenath <> writes:

> #include<stdio.h>
> char *f1(void);
> char *f1(void)
> {
> char *abc ="Hello";
>
> return abc;
> }
>
> int main(void )
> {
> char *p;
> p = f1();
>
> printf("%s\n", p);
> return 0;
> }
>
>
> While compiling the above program as mentioned below I am getting the
> meaning
> $ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
> Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
> prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
> Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
> jj.c
> jj.c: In function `f1':
> jj.c:5: warning: initialization discards qualifiers from pointer
> target type
>
> I have two question
> 1) What is the meaning of the warning ?

May it have anything to do with the signed-ness of char and the
signed-ness of "Hello"?

Asbjørn

Asbjørn wrote:
) somenath <> writes:
)
)> char *f1(void)
)> {
)> char *abc ="Hello";
)>
)> return abc;
)> }
)> <snip>
)> jj.c:5: warning: initialization discards qualifiers from pointer
)> target type
)>
)> I have two question
)> 1) What is the meaning of the warning ?
)
) May it have anything to do with the signed-ness of char and the
) signed-ness of "Hello"?

Unlikely. My first guess would rather be the const-ness.

About question 2): Is the above legal code ?

Yes it is. You're not returning the address of a local variable,
but that of a string literal. It's basically the same as:

char *f1(void)
{
return "Hello";
}

Which might give a similar warning, or might not.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

somenath said:

<snip>

> char *abc ="Hello";
>
<snip>
> jj.c:5: warning: initialization discards qualifiers from pointer
> target type
>
> I have two question
> 1) What is the meaning of the warning ?

There are, alas, no restrictions on the amount of gibberish that
implementations are allowed to produce when translating a program.

In this case, gcc is confusing "constant" with "const". The string literal,
"Hello", is an array of 6 char (so it has type char[6]), with static
storage duration, and it's "constant" in the sense that, if you try to
modify it, the behaviour is undefined. But it is *not* const. The
assignment doesn't discard any qualifiers at all.

gcc means well - it's trying to stop you from hurting yourself by pointing
an ordinary non-const-qualified char * to a string literal, which is
invariably a bad idea - but the wording of the warning is broken.


> 2) is it safe to return the local pointer value (i.e return abc is
> correct ?

It's unwise. Use const char * when pointing at string literals.

> My understanding is we should not return address of local
> variable .So above code may not be working always .

String literals are not variables, and their lifetime is not restricted to
that of the function (if any) in which they appear.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

somenath wrote:
>
> Hi All,
> I have one question regarding the code.
>
> #include<stdio.h>
> char *f1(void);
> char *f1(void)
> {
> char *abc ="Hello";
>
> return abc;
> }
>
> int main(void )
> {
> char *p;
> p = f1();
>
> printf("%s\n", p);
> return 0;
> }
>
> While compiling the above program as mentioned below I am getting the
> meaning
> $ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
> Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
> prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
> Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
> jj.c
> jj.c: In function `f1':
> jj.c:5: warning: initialization discards qualifiers from pointer
> target type
>
> I have two question
> 1) What is the meaning of the warning ?

Spurious.

> 2) is it safe to return the local pointer value (i.e return abc is
> correct ?
>
> My understanding is we should not return address of local
> variable .So above code may not be working always .

"abc" isn't a local pointer value.
"abc" has static duration. It exists from program start to end.

Your code is flawless. The warning is garbage.

I suspect that your compiler thinks that the type of ("abc" + 0)
is supposed to be (const char *),
but in C, the type of ("abc" + 0) is (char *).

In C,
the type of ("abc") is (char [4]),
not (const char [4]).

--
pete

somenath wrote:

> Hi All,
> I have one question regarding the code.
>
> #include<stdio.h>
> char *f1(void);
> char *f1(void)
> {
> char *abc ="Hello";
> return abc;
> }
>
> int main(void )
> {
> char *p;
> p = f1();
> printf("%s\n", p);
> return 0;
> }
>
>
> While compiling the above program as mentioned below I am getting the
> meaning
> $ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
> Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
> prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
> Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
> jj.c
> jj.c: In function `f1':
> jj.c:5: warning: initialization discards qualifiers from pointer
> target type
>
> I have two question
> 1) What is the meaning of the warning ?

Since you compiled with -Wwrite-strings, gcc non-Standardly makes
string literals be of type `const char *` rather than `char *`.
Then, when you try and assign the `const char *` to the `char *`
variable `abc`, it correctly complains: you've discarded the const
qualifier and left yourself open to someone changing something
which they're not supposed to change.

If you don't want this behaviour, don't ask for it.

> 2) is it safe to return the local pointer value (i.e return abc is
> correct ?

That isn't a "local pointer value"; it's the value of a local variable,
but that value is a pointer-to-string-literal, and a string literal
is a static and has lifetime equal to that of the entire executing
program.

> My understanding is we should not return address of local
> variable.

You shouldn't, and you aren't.

--
Chris "doctor, doctor, it hurts when I do /this/!" Dollin

Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN

pete wrote:
>
> somenath wrote:
> >
> > Hi All,
> > I have one question regarding the code.
> >
> > #include<stdio.h>
> > char *f1(void);
> > char *f1(void)
> > {
> > char *abc ="Hello";
> >
> > return abc;
> > }
> >
> > int main(void )
> > {
> > char *p;
> > p = f1();
> >
> > printf("%s\n", p);
> > return 0;
> > }
> >
> > While compiling the above program as mentioned below I am getting the
> > meaning
> > $ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
> > Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
> > prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
> > Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
> > jj.c
> > jj.c: In function `f1':
> > jj.c:5: warning: initialization discards qualifiers from pointer
> > target type
> >
> > I have two question
> > 1) What is the meaning of the warning ?
>
> Spurious.
>
> > 2) is it safe to return the local pointer value (i.e return abc is
> > correct ?
> >
> > My understanding is we should not return address of local
> > variable .So above code may not be working always .
>
> "abc" isn't a local pointer value.
> "abc" has static duration. It exists from program start to end.
>
> Your code is flawless. The warning is garbage.
>
> I suspect that your compiler thinks that the type of ("abc" + 0)
> is supposed to be (const char *),
> but in C, the type of ("abc" + 0) is (char *).
>
> In C,
> the type of ("abc") is (char [4]),
> not (const char [4]).

I forgot that we were talking about ("Hello") and not ("abc")

I suspect that your compiler thinks that the type of ("Hello" + 0)
is supposed to be (const char *),
but in C, the type of ("Hello" + 0) is (char *).

In C,
the type of ("Hello") is (char [6]),
not (const char [6]).

--
pete

somenath wrote:
>
> #include<stdio.h>
> char *f1(void);
> char *f1(void) {
> char *abc ="Hello";
> return abc;
> }
>
> int main(void ) {
> char *p;
> p = f1();
>
> printf("%s\n", p);
> return 0;
> }
>
> While compiling the above program as mentioned below I am getting
> the meaning
.... snip ...
> jj.c:5: warning: initialization discards qualifiers from pointer
> target type
>
> I have two question
> 1) What is the meaning of the warning ?
> 2) is it safe to return the local pointer value (i.e return abc
> is correct ?
>
> My understanding is we should not return address of local
> variable .So above code may not be working always .

Change "char *abc ="Hello";" to "const char *abc ="Hello";". You
are not returning a pointer to local storage, you are returning the
value of that local storage.

--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.



--
Posted via a free Usenet account from http://www.teranews.com

In article <>,
Richard Heathfield <> wrote:

>In this case, gcc is confusing "constant" with "const".

.... which the user asked it to do, by means of the flag -Wwrite-strings:

`-Wwrite-strings'
When compiling C, give string constants the type `const
char[LENGTH]' so that copying the address of one into a
non-`const' `char *' pointer will get a warning; when compiling
C++, warn about the deprecated conversion from string constants to
`char *'. These warnings will help you find at compile time code
that can try to write into a string constant, but only if you have
been very careful about using `const' in declarations and
prototypes. Otherwise, it will just be a nuisance; this is why we
did not make `-Wall' request these warnings.

-- Richard
--
:wq

pete wrote:
> pete wrote:
>> somenath wrote:
>>
.... snip ...
>>
>>> $ gcc -W -Wall -ansi -pedantic -Wformat-nonliteral -Wcast-align -
>>> Wpointer-arith -Wbad-function-cast -Wmissing-prototypes -Wstrict-
>>> prototypes -Wmissing-declarations -Winline -Wundef -Wnested-externs -
>>> Wcast-qual -Wshadow -Wconversion -Wwrite-strings -ffloat-store -O2
^^^^^^^^^^^^^^^
.... snip ...
>>
>> Spurious.

Not spurious.

.... snip ...
>>
>> Your code is flawless. The warning is garbage.

The code is flawed. The warnings are accurate.

>> I suspect that your compiler thinks that the type of ("abc" + 0)
>> is supposed to be (const char *), but in C, the type of
>> ("abc" + 0) is (char *).

Look up the meaning of the underlined gcc option above.

.... snip ...
>
> In C, the type of ("Hello") is (char [6]), not (const char [6]).

Not when it has been told otherwise.

--
Chuck F (cbfalconer at maineline dot net)
<http://cbfalconer.home.att.net>
Try the download section.



--
Posted via a free Usenet account from http://www.teranews.com