Typecasting in C

[andynaik]

Hi,
Whenever we type in this code
int main()
{
printf("%f",10) ;
}
we get an error. We can remove that by using #pragma directive t
direct that to the 8087. Even after that the output is 0.00000 and no
10.0000. Can anybody tell me why it is like that and why typecasting i
not done in this case?
-
andynai
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[Mark McIntyre]

On Tue, 29 Jun 2004 07:56:37 GMT, in comp.lang.c , andynaik
<an************ *@mail.codecomm ents.com> wrote:

Hi,
Whenever we type in this code
int main()
{
printf("%f",10 );
}
we get an error.
Yes, because the code is broken.
We can remove that by using #pragma directive to
direct that to the 8087.
Don't do that, you're breaking it even worse.
Even after that the output is 0.00000 and not
10.0000. Can anybody tell me why it is like that and why typecasting is
not done in this case??

This has nothing to do with typecasting. You're calling printf() without
any prototype or declaration in scope, so it has no idea what the arguments
mean. You absolutely must declare functions before you use them, typically
by #including the correct header.
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.c om/ms3/bchambless0/welcome_to_clc. html>
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[Keith Thompson]

Mark McIntyre <ma**********@s pamcop.net> writes:

On Tue, 29 Jun 2004 07:56:37 GMT, in comp.lang.c , andynaik
<an************ *@mail.codecomm ents.com> wrote:

Whenever we type in this code
int main()
{
printf("%f",10 );
}
we get an error.
Yes, because the code is broken.

[...] This has nothing to do with typecasting. You're calling printf() without
any prototype or declaration in scope, so it has no idea what the arguments
mean. You absolutely must declare functions before you use them, typically
by #including the correct header.

Including the <stdio.h> header to declare printf() won't solve this
problem (though of course you should do it anyway). For a function
that takes a variable number of arguments, the compiler still doesn't
know the types of the variable arguments. You have to make sure the
types of the arguments match the expected types for the format.

Some compilers may issue warnings based on the format string (if it's
a literal), but they won't do the conversions for you.

(BTW, Mark, your newsfeed.com sig is appearing twice on each post.)

--
Keith Thompson (The_Other_Keit h) 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.

[Emmanuel Delahaye]

In 'comp.lang.c', "jacob navia" <ja***@jacob.re mcomp.fr> wrote:

printf("the address is: 0x%x\n",p);

where p is some pointer appears in several million lines in
existing code.

But it's wrong. On a x86 platform, it will fail in large memory model (at
least).

#include <stdio.h>

int main (void)
{
int a;
int *p = &a;

printf ("addr = %x\n", p);
printf ("addr = %p\n", (void *) p);
return 0;
}

Memory model small

D:\CLC\E\EMMANU EL>bc proj.prj
addr = fff4
addr = FFF4

Memory model large

D:\CLC\E\EMMANU EL>bc proj.prj
addr = ffe
addr = 8FF0:0FFE
--
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[Emmanuel Delahaye]

In 'comp.lang.c', "Arthur J. O'Dwyer" <aj*@nospam.and rew.cmu.edu> wrote:

Who's been using "%d" or "%x" to print *pointer* values?

I probably did in the earliest times, but not since I have used BC 3.1
(1991). I have learnt about the (void*) thing by reading c.l.c since 1999.

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