how to convert a program to a function/macro and put it in a header
file? is there any shortcut method for this?
thanks.
file? is there any shortcut method for this?
thanks.
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Re: Function & header files
Umesh said:
Don't even try. Instead, find out how to use libraries.how to convert a program to a function/macro and put it in a header
file?
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at the above domain, - www.
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Re: Function & header files
On 12 Jun, 14:05, Richard Heathfield <r...@see.sig.i nvalidwrote:Now you've done it. If he picks up on this (he may not, as heUmesh said:
>>how to convert a program to a function/macro and put it in a header
file?
Don't even try. Instead, find out how to use libraries.
generally wants to be spoonfed a direct answer to his question), he'll
now ask "how to use libraries?" and we'll have a heap of "off-topic"
responses...
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function -> macro
How can I convert a function to a macro in general? e.g.
void factorial(int n)
{
long int x=1;
for (int i=1;i<=n;i++)
x*=i;
printf("Factori al of %d = %ld\n",n,x);
}
Comment
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Re: function -> macro
Umesh wrote:
| How can I convert a function to a macro in general? e.g.
|
| void factorial(int n)
| {
| long int x=1;
| for (int i=1;i<=n;i++)
| x*=i;
| printf("Factori al of %d = %ld\n",n,x);
| }
Sorry - there is no "general solution". Please re-read Richard's
article.
--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
Comment
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Re: function -> macro
Umesh said:
You'd be better off asking "how can I avoid risking the generation ofHow can I convert a function to a macro in general? e.g.
>
void factorial(int n)
{
long int x=1;
for (int i=1;i<=n;i++)
x*=i;
printf("Factori al of %d = %ld\n",n,x);
}
incorrect results?"
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at the above domain, - www.
Comment
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Re: function -> macro
Richard Heathfield <rjh@see.sig.in validwrites:He'd be *far* better off asking "How do I use this Usenet thingUmesh said:>>How can I convert a function to a macro in general? e.g.
>>
>void factorial(int n)
>{
> long int x=1;
> for (int i=1;i<=n;i++)
> x*=i;
> printf("Factori al of %d = %ld\n",n,x);
>}
You'd be better off asking "how can I avoid risking the generation of
incorrect results?"
without looking like a parasite?".
--
Keith Thompson (The_Other_Keit h) kst-u@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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Comment
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Re: function -> macro
Umesh wrote:A function that declares local variables,>
How can I convert a function to a macro in general? e.g.
>
void factorial(int n)
{
long int x=1;
for (int i=1;i<=n;i++)
x*=i;
printf("Factori al of %d = %ld\n",n,x);
}
is a poor candidate for conversion to a macro.
/* BEGIN new.c */
#include <stdio.h>
#define factorial(N) \
do { \
n = N; \
x = i = 1; \
while (n >= i) { \
x *= i++; \
} \
printf("Factori al of %d = %ld\n", n, x); \
} while (0)
void (factorial)(int n);
int main(void)
{
long int x = 1, i = 1, n;
factorial(5);
(*factorial)(5) ;
return 0;
}
void (factorial)(int n)
{
long int x = 1, i = 1;
factorial(n);
}
/* END new.c */
--
pete
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Re: function -> macro
// factorial by macro
#include<stdio. h>
#define factorial(n) \
long int x=1; \
for(int i=1;i<=n;i++) \
x*=i; \
printf("Factori al of %d = %ld\n",n,x);\
main()
{
int n;
printf("Enter No.- ");
scanf("%d",&n);
factorial(n);
return 0;
}
Comment
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Re: function -> macro
Umesh said:
foo.c:9: warning: return-type defaults to `int'// factorial by macro
>
#include<stdio. h>
#define factorial(n) \
long int x=1; \
for(int i=1;i<=n;i++) \
x*=i; \
printf("Factori al of %d = %ld\n",n,x);\
>
main()
{
int n;
printf("Enter No.- ");
scanf("%d",&n);
factorial(n);
return 0;
}
foo.c:9: warning: function declaration isn't a prototype
foo.c: In function `main':
foo.c:13: parse error before `long'
foo.c:13: parse error before `int'
foo.c:13: `i' undeclared (first use in this function)
foo.c:13: (Each undeclared identifier is reported only once
foo.c:13: for each function it appears in.)
foo.c:13: warning: statement with no effect
foo.c:13: parse error before `)'
foo.c:13: `x' undeclared (first use in this function)
make: *** [foo.o] Error 1
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at the above domain, - www.
Comment
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Re: function -> macro
Richard Heathfield <rjh@see.sig.in validwrites:[snip]Umesh said:
>>>// factorial by macro
>>
>#include<stdio .h>
>#define factorial(n) \
>long int x=1; \
>for(int i=1;i<=n;i++) \
>x*=i; \
>printf("Factor ial of %d = %ld\n",n,x);\
>>
>main()
>{
>int n;
>printf("Ente r No.- ");
>scanf("%d",&n) ;
>factorial(n) ;
>return 0;
>}
foo.c:9: warning: return-type defaults to `int'
foo.c:9: warning: function declaration isn't a prototype
foo.c: In function `main':
foo.c:13: parse error before `long'
foo.c:13: parse error before `int'
Umesh is obviously using a compiler that implements some C99 features.
Using such features isn't wrong, merely non-portable. When I compile
with "gcc -std=c99 -pedantic -Wall -W -O3", the only diagnostic I get
is "warning: return type defaults to `int'".
Of course, the program is horribly bad for other reasons. The use of
a macro is entirely gratuitous, and the macro unaccountably prints the
result rather than yielding it as a result.
Here's another program that uses a macro to compute factorials:
#include <stdio.h>
static unsigned long factorial_func( unsigned n)
{
unsigned long f = 1;
unsigned i;
for (i = 2; i <= n; i ++) {
f *= i;
}
return f;
}
#define factorial(n) factorial_func( n)
int main(void)
{
unsigned n;
printf("Enter number: ");
scanf("%u", &n);
printf("factori al(%u) = %lu\n", n, factorial(n));
return 0;
}
The use of a macro is equally gratuitous, but at least this one
operates more usefully. On a system with 32-bit longs, either program
works only for arguments up to 12; using unsigned long rather than
long didn't add enough range to change that. With 64-bit longs, it
works for arguments up to 20. Neither version checks for overflow
(which would be fairly difficult to do), but my version uses unsigned
arithmetic and therefore avoids undefined behavior. The use of scanf
for input is dangerous, as I've discussed here before, but I was too
lazy to correct it.
--
Keith Thompson (The_Other_Keit h) kst-u@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."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Comment
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Re: function -> macro
Which compiler do you use?
It's working fine in TC++ 3/4.5 & VC++ 6
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Re: function -> macro
Umesh said:
When writing C programs, I am perfectly content to use *any* compilerWhich compiler do you use?
that supports either ISO/IEC 9899:1990 when invoked in conforming mode
or ISO/IEC 9899:1999 when invoked in conforming mode.
Neither of those compilers claims to conform to ISO/IEC 9899:1999. TheyIt's working fine in TC++ 3/4.5 & VC++ 6
do, however, claim to conform to ISO/IEC 9899:1990, so that's the
standard by which programs written for them should be judged. By that
standard, your program is invalid for the reasons stated.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at the above domain, - www.
Comment
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Re: function -> macro
WHICH COMPILER(S) DO YOU USE?
GCC and other C compilers now support many of the new features of C99.
However, there has been less support from vendors such as Microsoft
and Borland that have mainly focused on C++, since C++ provides
similar functionality improvement.
GCC, despite its extensive C99 support, is still not a completely
compliant implementation; several key features are missing or don't
work correctly
Richard Heathfield wrote:
Umesh said:
>>Which compiler do you use?
When writing C programs, I am perfectly content to use *any* compiler
that supports either ISO/IEC 9899:1990 when invoked in conforming mode
or ISO/IEC 9899:1999 when invoked in conforming mode.
>>It's working fine in TC++ 3/4.5 & VC++ 6
Neither of those compilers claims to conform to ISO/IEC 9899:1999. They
do, however, claim to conform to ISO/IEC 9899:1990, so that's the
standard by which programs written for them should be judged. By that
standard, your program is invalid for the reasons stated.
>
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at the above domain, - www.Comment
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Re: function -> macro
"Umesh" <fraternitydisp osal@gmail.comh a scritto nel messaggio
news:1181912917 .184583.62090@z 28g2000prd.goog legroups.com...Well, I see that at least you know how to copy and paste.WHICH COMPILER(S) DO YOU USE?
>
GCC and other C compilers now support many of the new features of C99.
However, there has been less support from vendors such as Microsoft
and Borland that have mainly focused on C++, since C++ provides
similar functionality improvement.
>
GCC, despite its extensive C99 support, is still not a completely
compliant implementation; several key features are missing or don't
work correctly
>
Richard Heathfield wrote:
>>>Umesh said:
>>>>Which compiler do you use?
>When writing C programs, I am perfectly content to use *any* compiler
>that supports either ISO/IEC 9899:1990 when invoked in conforming mode
>or ISO/IEC 9899:1999 when invoked in conforming mode.
>>>>It's working fine in TC++ 3/4.5 & VC++ 6
>Neither of those compilers claims to conform to ISO/IEC 9899:1999. They
>do, however, claim to conform to ISO/IEC 9899:1990, so that's the
>standard by which programs written for them should be judged. By that
>standard, your program is invalid for the reasons stated.
>>
>--
>Richard Heathfield
>"Usenet is a strange place" - dmr 29/7/1999
>http://www.cpax.org.uk
>email: rjh at the above domain, - www.
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