Abridged FAQ Question

[dave]

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

A: The first declares an initialized and modifiable array; the
second declares a pointer initialized to a not-necessarily-
modifiable constant string.
I have this code in gcc (yes, it's C++, but it is a C question):

char * char_ptr = "stringthingy";

cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n";
cout << "*char_ptr = " << *char_ptr << "\n";
cout << "char_ptr = " << char_ptr << "\n";

And this is the result:

Size of char_ptr is 4

*char_ptr = s
char_ptr = stringthingy

So sizeof() returns 4 as expected and the dereferenced pointer
returns 's' as expected. But why isn't an address returned for
the non-dereferenced pointer? If "stringthingy" is correct, why
doesn't sizeof() return 13?

I am confused!
TIA,

~Dave~

[Keith Thompson]

dave <da**@comteck.com> writes:

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

A: The first declares an initialized and modifiable array; the
second declares a pointer initialized to a not-necessarily-
modifiable constant string.
I have this code in gcc (yes, it's C++, but it is a C question):

char * char_ptr = "stringthingy";

cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n";
cout << "*char_ptr = " << *char_ptr << "\n";
cout << "char_ptr = " << char_ptr << "\n";

And this is the result:

Size of char_ptr is 4

*char_ptr = s
char_ptr = stringthingy

So sizeof() returns 4 as expected and the dereferenced pointer
returns 's' as expected. But why isn't an address returned for
the non-dereferenced pointer? If "stringthingy" is correct, why
doesn't sizeof() return 13?

It would have been much better to post C code. The behavior of the
C++ code you posted depends on the way C++ operator overload works.
It's also helpful to post a complete program rather than a code
fragment.

Here's a C program that corresponds to the C++ code you posted:

#include <stdio.h>
int main(void)
{
char *char_ptr = "stringthingy";

printf("Size of char_ptr is %d\n\n", sizeof char_ptr);
printf("*char_ptr = %c\n", *char_ptr);
printf("char_ptr = %s\n", char_ptr);
return 0;
}

In the last printf() call, the "%s" format expects a char* value that
points to a string, and that's what you give it. The pointer is
dererenced inside printf() itself.

If you wanted to print the pointer value itself, you could use the
"%p" format, which expects a void*:

printf("char_ptr = %p\n", (void*)char_ptr);

(The cast to void* isn't strictly necessary in this case for obscure
reasons, but it's a good idea.)

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Joe Wright]

dave wrote:

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

A: The first declares an initialized and modifiable array; the
second declares a pointer initialized to a not-necessarily-
modifiable constant string.
I'm going to ignore the C++ code below. It is not topical. Perhaps the
Answer above might be confusing you. Let me try.

char a[] = "string";

...describes an array of char in memory, the first character of which is
at address a. Now..

char *p = "string";

...describes a pointer which holds the address of the anonymous string.
This is two objects, the pointer and the string. In the first case we
have only one object, the array a.


I have this code in gcc (yes, it's C++, but it is a C question):

char * char_ptr = "stringthingy";

cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n";
cout << "*char_ptr = " << *char_ptr << "\n";
cout << "char_ptr = " << char_ptr << "\n";

And this is the result:

Size of char_ptr is 4

*char_ptr = s
char_ptr = stringthingy

So sizeof() returns 4 as expected and the dereferenced pointer
returns 's' as expected. But why isn't an address returned for
the non-dereferenced pointer? If "stringthingy" is correct, why
doesn't sizeof() return 13?

I am confused!
TIA,

~Dave~

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[Jack Klein]

On Mon, 06 Feb 2006 09:25:23 GMT, Keith Thompson <ks***@mib.org> wrote
in comp.lang.c:

dave <da**@comteck.com> writes:

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

A: The first declares an initialized and modifiable array; the
second declares a pointer initialized to a not-necessarily-
modifiable constant string.
I have this code in gcc (yes, it's C++, but it is a C question):

char * char_ptr = "stringthingy";

cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n";
cout << "*char_ptr = " << *char_ptr << "\n";
cout << "char_ptr = " << char_ptr << "\n";

And this is the result:

Size of char_ptr is 4

*char_ptr = s
char_ptr = stringthingy

So sizeof() returns 4 as expected and the dereferenced pointer
returns 's' as expected. But why isn't an address returned for
the non-dereferenced pointer? If "stringthingy" is correct, why
doesn't sizeof() return 13?
It would have been much better to post C code. The behavior of the
C++ code you posted depends on the way C++ operator overload works.
It's also helpful to post a complete program rather than a code
fragment.

Here's a C program that corresponds to the C++ code you posted:

#include <stdio.h>
int main(void)
{
char *char_ptr = "stringthingy";

printf("Size of char_ptr is %d\n\n", sizeof char_ptr);

ITYM:
printf("Size of char_ptr is %d\n\n", (int)sizeof char_ptr);
printf("*char_ptr = %c\n", *char_ptr);
printf("char_ptr = %s\n", char_ptr);
return 0;
}

In the last printf() call, the "%s" format expects a char* value that
points to a string, and that's what you give it. The pointer is
dererenced inside printf() itself.

If you wanted to print the pointer value itself, you could use the
"%p" format, which expects a void*:

printf("char_ptr = %p\n", (void*)char_ptr);

(The cast to void* isn't strictly necessary in this case for obscure
reasons, but it's a good idea.)

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Keith Thompson]

Jack Klein <ja*******@spamcop.net> writes:

On Mon, 06 Feb 2006 09:25:23 GMT, Keith Thompson <ks***@mib.org> wrote
in comp.lang.c:

[...]

printf("Size of char_ptr is %d\n\n", sizeof char_ptr);

ITYM:
printf("Size of char_ptr is %d\n\n", (int)sizeof char_ptr);

Yes, thanks.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[fangshi]

If you want to get the address that p point to you can do like this:
printf("%d",p);
If printf(p) or printf("%s",p) ,we also get string value ,not its
address.
I don't know exactly why isn't an address returned.

dave 写�：

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

A: The first declares an initialized and modifiable array; the
second declares a pointer initialized to a not-necessarily-
modifiable constant string.
I have this code in gcc (yes, it's C++, but it is a C question):

char * char_ptr = "stringthingy";

cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n";
cout << "*char_ptr = " << *char_ptr << "\n";
cout << "char_ptr = " << char_ptr << "\n";

And this is the result:

Size of char_ptr is 4

*char_ptr = s
char_ptr = stringthingy

So sizeof() returns 4 as expected and the dereferenced pointer
returns 's' as expected. But why isn't an address returned for
the non-dereferenced pointer? If "stringthingy" is correct, why
doesn't sizeof() return 13?

I am confused!


TIA,

~Dave~

[Keith Thompson]

"fangshi" <fa***********@gmail.com> writes:

If you want to get the address that p point to you can do like this:
printf("%d",p);
If printf(p) or printf("%s",p) ,we also get string value ,not its
address.
I don't know exactly why isn't an address returned.

Please don't top-post. See <http://www.caliburn.nl/topposting.html>
for more information.

If p is a pointer, such as
char *p = "string";
then
printf("%d", p);
*might* "work" on some implementations, but it's dangerously
non-portable (it invokes undefined behavio). Use the "%p" format
to print a pointer value:
printf("%p", (void*)p);

printf(p) can be dangerous, since the string pointed to by p might
contain format specifiers. printf("%s", p) is much safer. (As I
mentioned in another response in this thread, it prints the string
because the pointer is dereferenced inside the printf() function.)

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Flash Gordon]

fangshi wrote:

Please don't top post. You response belongs after or intermixed with the
text you are replying to.

Top posting fixed.

From the FAQ:

1.32: What is the difference between char a[] = "string"; and
char *p = "string"; ?

<snip>
If you want to get the address that p point to you can do like this:
printf("%d",p);
No you can't, at least not if you want code that is guaranteed to work.
Your suggestion will fail on real systems. The correct way to print the
value of a pointer is
printf("%p",(void*)p);
If printf(p) or printf("%s",p) ,we also get string value ,not its
address.
Your first suggestion [printf(p)] is, in general, a very bad idea. If p
points to the string "99% rubbish" you have just invoked undefined
behaviour and anything can happen, the most likely result being it not
printing what you expected.
I don't know exactly why isn't an address returned.

Do you mean why an address isn't printed? If so, the answer is you have
not told it to print an address.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.