hello everyone,
i was wondering i know that with scanf() you are supposed to pass it the
address in memory where you want the value assigned like if it was a variable:

int a;
[snip]
scanf("%d", &a);

and if it where all ready a pointer:

int *a;
[snip]
scanf("%d", a);

but what about printf()?
normally you just pass it the name of the variable:

int a;
[snip]
printf("%d", a);

so does that mean when working with a you do:

int *a;
[snip]
printf("%d", *a);

or:

int *a;
[snip]
printf("%d", a);

i have heard that scanf() is just the complete opposite of printf() but i did
not take it literally! does this mean that you pass printf() the actual value
and that you pass scanf() the address in memory? are they really truely
opposite like that?

note: if they really are opposite like that thsi would be the right one:

int *a;
[snip]
printf("%d", *a);

but i do not know

thank you everyone and i hope my question was not so stupid

-art  

This gets an integer into a...

...and this gets an integer into the int that a points to.  Note that
that's not exactly the same thing, though in each case what your code
snippets do is usually The Right Thing.

This prints the int that is stored in a...

...And this prints the int that a points to, which is the equivalent of
the scanf snippet that uses a pointer above.

This is incorrect; if you want to print the integer value, you'd want
to pass *a as above.  It's also possible to print a pointer itself:
printf("%p",(void *)a);
This prints a pointer value, which on many systems will look something
like '0xdeadbeef' (which is a hexadecimal value representing the address
that the pointer points to).  (This probably isn't what you were asking
about, but it's interesting and sometimes even useful.)

They're not exact opposites, since some conversion characters (such as %i)
act differently for scanf than the opposite of printf, but in the case
of whether to pass pointers or values it's probably easiest to think of
them as opposites.  (Of course, it's probably even better to not bother
with scanf at all and use other, more robust ways of getting input.)

dave

--

So would you say that this is sickly contorting, maximally probable code?
                                             --Kaz Kylheku in comp.lang.c

perfect.

no.

Its not - scanf must take pointers to its arguments, so that it can
store data in them and return the modified data. printf only needs the
args themselves, so it can read from them
but i did

--
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>

thanks for your help!

this make perfect sense to me except for one part:

(void *)

i gather you can put any data type such as (char *) or (int *)
but hwat does this do and why is it needed? if you could explain i think it
would help me alot? my hypotheses for right now is that it multiplies that data
type by itself and the value ends up being that like:

(float *) would evaluate to:

4

or

(double *) to:

8

thanks and please tell me if my guess is wrong!

%p requires a pointer. void* is a generic pointer. So you're casting a
to a pointer.

Not sure what that means !

No, it will cast the value to a pointer type, and hten printf will
print out hte value of that pointer.

--
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>

In this case, a was already a pointer (to int), but %p specifically
requires a pointer to void, and since printf is a variadic function[1]
the conversion to (void *) needs to be done explicitly.

In general, for a type T, (T)foo tells the compiler you want to convert
foo to type T.  Often this is done automatically, but in some cases
(like passing arguments to variadic functions) it isn't so you need
to specifically tell the compiler to do it.  Casts can also be used
to tell the compiler ``Yes, I really want to do this, and know exactly
what I'm doing'' for a conversion that it would otherwise warn about;
doing this is usually a Bad Thing, and needing a cast to silence a
compiler warning is usually a sign that you're doing something wrong.

[1] A variadic function is one that can take varying numbers of arguments;
    the relevance here is that the default argument promotions are applied
    to arguments after the last explicitly specified one (e.g. a short
    is converted to an int), and the default promotions don't convert
    pointers (a pointer to <foo> is always passed as a pointer to <foo>).

dave

--

Absolutely right. Actually telling people they're in the wrong newsgroup for
their question would contravene their constitutional right to be ignorant.
                          --Richard Heathfield roasts a troll in comp.lang.c

ok so say i did this:

int *a;
[snip]
printf("%c, (char *)*a);

this would change the data type of the value a points to to a char data type?(i
know that this is not needed i did this just for example purposes) and does it
change it to the char data type for the rest of the program or just for that
function?

thanks!

-art

    What (char *) *a is going to do is take the value pointed to by a and treat it
as though it were a pointer to a character.  The actual data type doesn't get
changed.  Only the way it is treated at that particular time.

Christopher Lansing

No, it would only convert the value of the int that a points to to
a pointer to char.  (Casts work the same way as other operators;
a+1 would give you a pointer to the next int past the one a points to
without changing a.)  It would then pass the pointer to char to printf.
You probably don't want to cast an int to a pointer, especially not here
since %c expects a char; what you're looking for here is
printf("%c",(char)*a);

A cast returns the value that is the result of converting the value
you're casting to whatever type you're casting it to; it doesn't change
any data types, it only converts values.

dave

--

Absolutely right. Actually telling people they're in the wrong newsgroup for
their question would contravene their constitutional right to be ignorant.
                          --Richard Heathfield roasts a troll in comp.lang.c

I wondered if that might be the case, but too much context was snipped
unfortunately.

--
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>

GLOBAL LINK 2001 a crit dans le message

If you insist to use scanf() (which is really a bad idea, as you fell not to
understand), at last do it proberly, and check the return value.

Wrong. The pointer must point to something valid.

int b;
int *a = &b;
   scanf("%d", a);

Yes, assuming a holds a valid value and not a trap representation. Also, for
portability reasons, better to finish lines with '\n'

   printf("%d\n", a);

Horribly wrong. Neither a nor *a are initialized.

Horribly wrong again. a is still not initialized. If you want to printf() a
valid pointer, use "%p" and the (void*) cast.

int b;
int *a = &b;
   printf("%p\n", (void*)a);

did

You heard badly. Don't learn by rumors. Better to understand the C-language.

Just read a good C-reference, and comply with the definition of the
functions:
http://www.dinkum.com/htm_cl/lib_prin.html
http://www.dinkum.com/htm_cl/lib_scan.html

What opposite? They are different. Period.

You should know that dereferencing a non-initialized pointer is a Bad Thing.

--
-hs-    Tabs out, spaces in.
CLC-FAQ: http://www.eskimo.com/~scs/C-faq/top.html
ISO-C Library: http://www.dinkum.com/htm_cl
FAQ de FCLC : http://www.isty-info.uvsq.fr/~rumeau/fclc

I do not understand this advice. Portability with what?
Please elaborate.

willem

A text stream is composted of lines, a line is terminated by '\n'.

It is implementation-defined if the last line need a terminating '\n' or
not.

--
Tor <torust AT online DOT no>

From 7.19.3:

When a stream is line buffered, characters are intended to be
transmitted to or from the host environment as a block when a new-line
character is encountered.

If you want to guarantee that your prompt will appear before your
program starts blocking for input, the newline (or an fflush) is a good
plan.

--
Richard Heathfield
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

printf("\nHey, Richard");
printf("\nSee, I can count to ten too:");
for (i = 0; i < 10; i++) printf("%3d", i);
printf("\nI do not always want \n, because that breaks up your line");
printf("\nSo what is wrong with this syntax?"
       "\nEven though it has no linefeeds at the end ..");
printf("\nwillem (you'll even see this when I exit or even sooner :)";