function problem

[Bill Cunningham]

I know this must have some simple answer but I'm having trouble with this
function I wrote.
stdio.h
math.h

double relstr (double x,double y)
{double n;
n=x/y;
printf("%f",(return n;));}
I dunno.

Bill

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[Morris Dovey]

Bill Cunningham wrote:

I know this must have some simple answer but I'm having trouble with this
function I wrote.
stdio.h
math.h

double relstr (double x,double y)
{double n;
n=x/y;
printf("%f",(return n;));}
I dunno.

Bill...

I hope your C compiler is having trouble with this one, too.
You're trying to code a statement where the only allowable
construct is an expression.

--
Morris Dovey
West Des Moines, Iowa USA
C links at http://www.iedu.com/c
Read my lips: The apple doesn't fall far from the tree.

[Jack Klein]

On Thu, 15 Jan 2004 21:14:28 -0500, "Bill Cunningham"
<no****@nspam.net> wrote in comp.lang.c:

I know this must have some simple answer but I'm having trouble with this
function I wrote.
stdio.h
math.h

double relstr (double x,double y)
{double n;
n=x/y;
printf("%f",(return n;));}
I dunno.

Bill

One of your problems is that you are using a broken compiler, if it
compiles this, since the code is illegal. You can't use a return
statement as an argument to a function, as it does not yield a value.

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

[Bill Cunningham]

One of your problems is that you are using a broken compiler, if it
compiles this, since the code is illegal. You can't use a return
statement as an argument to a function, as it does not yield a value.

Oh no the compiler won't compile this. It says error before return. Don't I
have to get the code to return the value of n and send it to stdout via
printf?

Bill

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[pete]

Bill Cunningham wrote:

One of your problems is that you are using a broken compiler, if it
compiles this, since the code is illegal. You can't use a return
statement as an argument to a function,
as it does not yield a value.

Oh no the compiler won't compile this. It says error before return.
Don't I have to get the code to return the value of n and
send it to stdout via printf?

#include <stdio.h>

double relstr (double x, double y)
{
double n;

n = x / y;
printf("%f", n);
return n;
}

--
pete

[Mark McIntyre]

On Thu, 15 Jan 2004 22:15:32 -0500, in comp.lang.c , "Bill Cunningham"
<no****@nspam.net> wrote:

One of your problems is that you are using a broken compiler, if it
compiles this, since the code is illegal. You can't use a return
statement as an argument to a function, as it does not yield a value.
Oh no the compiler won't compile this

thats a relief.
. It says error before return. Don't I
have to get the code to return the value of n and send it to stdout via
printf?

read what Jack said again.,
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.com/ms3/bchambless0/welcome_to_clc.html>
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[Mark McIntyre]

On Thu, 15 Jan 2004 21:14:28 -0500, in comp.lang.c , "Bill Cunningham"
<no****@nspam.net> wrote:

I know this must have some simple answer but I'm having trouble with this
function I wrote. stdio.h
math.h
first off, get a better newsreader - one that doesn't strip out the
#include and <>'s from your posts !!
double relstr (double x,double y)
{double n;
n=x/y;
printf("%f",(return n;));}

Two problems here:
1) you can't use a return statement as an argument of a function.
2) your brace style is EXCEPTIONALLY horrible. Do NOT follow or
precede braces with code, it makes it unbelievable hard to read.

--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.com/ms3/bchambless0/welcome_to_clc.html>
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[Old Wolf]

> double relstr (double x, double y)

{
double n;

n = x / y;
printf("%f", n);

ITYM:
printf("%lf", n);

[pete]

Old Wolf wrote:

double relstr (double x, double y)
{
double n;

n = x / y;
printf("%f", n);

ITYM:
printf("%lf", n);

No.

--
pete

[Chris Torek]

>> printf("%f", n);
[where n is a "double"]

In article <news:84**************************@posting.google. com>
Old Wolf <ol*****@inspire.net.nz> writs:

ITYM:
printf("%lf", n);

This is well-defined in C99, and does the same as using "%f", but
in C89 the combination of the "l" modifier with any of the floating
point formats in printf (e, E, f, g, and G) is technically undefined.

You always need the l in %lf when using scanf() on "double"s, but
in C89, you should not use the l modifier. The printf() and scanf()
functions are not as symmetric as they might first appear. This
false symmetry begins with the fact that scanf() takes the addresses
of the objects it fills in, and continues with %f vs %lf and gets
even more asymmetric with the behavior of %* directives. In printf,
for instance, %*d requires two int values and prints the second
using the first as a field width, but in scanf, %*d requires zero
pointers and skips the assignment of an "int" conversion.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.