Why does math.h not declare function round?

Why does math.h not declare function round?

[Plato]

Hi!
    I made such a program foo.c:

#include <math.h>

int main()
{
    double bar=1.2;
    printf("%f\n", round(bar));
}
    
Then compile and run:
[root@so41 tmp]# gcc foo.c -ofoo -lm
[root@so41 tmp]# ./foo
0.000000

The result was obviously wrong. Then I add statement:
#include <math.h>
double round(double x);

int main()
{
    double bar=1.2;
    printf("%f\n", round(bar));
}

Then compile and run:
[root@so41 tmp]# gcc foo.c -ofoo -lm
[root@so41 tmp]# ./foo
1.000000

Now it's right.

I wonder why function round not declared in math.h. Anyone have idea?

Thx.




Plato
2007-09-24

Re: Why does math.h not declare function round?

[Steve Graegert]

On 9/24/07, Plato <zhangguoqiang@xunlei.com> wrote:
> Hi!
>     I made such a program foo.c:
>
> #include <math.h>
>
> int main()
> {
>     double bar=1.2;
>     printf("%f\n", round(bar));
> }
>
> Then compile and run:
> [root@so41 tmp]# gcc foo.c -ofoo -lm
> [root@so41 tmp]# ./foo
> 0.000000
>
> The result was obviously wrong. Then I add statement:
> #include <math.h>
> double round(double x);
>
> int main()
> {
>     double bar=1.2;
>     printf("%f\n", round(bar));
> }
>
> Then compile and run:
> [root@so41 tmp]# gcc foo.c -ofoo -lm
> [root@so41 tmp]# ./foo
> 1.000000
>
> Now it's right.
>
> I wonder why function round not declared in math.h. Anyone have idea?

It is available in C99 mode only (see <bits/mathcalls.h>):

#ifdef __USE_ISOC99
__MATHCALLX (round,, (_Mdouble_ __x), (__const__));

To write fully portable code use floor() and ceil() instead.

	\Steve

--

Steve Grägert <steve@graegert.com>
Internet: http://digitalether.de
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