6

Is there a good reason why this program compiles under GCC even with the -ansi and -pedantic flags?

#include <cmath>

int main (int argc, char *argv [])
{
     double x = 0.5;

     return static_cast<int>(round(x));
}

This compiles clean (no warnings, even) with g++ -ansi -pedantic -Wall test.cpp -o test.

I see two problems:

  1. round() shouldn't be available to C++ in ISO-conformant mode (since it comes from C99)
  2. Even if round() were available in this case, it should only be so from the std namespace

Am I wrong?

Antti Haapala -- Слава Україні
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Dan Moulding
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3 Answers3

2

This is a bug. It's been around for a surprisingly long while. Apparently, there has not been enough of a collective desire to fix it. With a new version of C++ just around the corner which will adopt the C99 functions from math.h, it seems unlikely it will ever be fixed.

Dan Moulding
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0

I might be off base here but doesn't gcc's -ansi flag apply to the code constructs (ie, disable GCC language extensions) rather than switching all libraries into strict ANSI compliant mode as well?

Timo Geusch
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  • the man page to gcc states that some function will not be defined when the ansi flag is set, but not that only ansi functions will be defined – josefx Dec 10 '09 at 17:54
  • @josefx: See my comments to D. Shawley. After investigating more, I've learned that library extensions cannot "alter the behavior of well-formed programs". So it seems to me that adding a `round()` extension to the library would be illegal, which makes the behavior of the `-ansi` flag moot. – Dan Moulding Dec 10 '09 at 19:04
-1

I believe that the Standards specify what symbols are required to be defined and in which header they are defined. I do not believe that the Standards state that no other symbols may be defined. More to the point, std::round() will not be defined by a free symbol called round() can be defined.

D.Shawley
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  • Okay, that may be true but the purpose of `-ansi`, according to the GCC manual, is to *disable* all non-standard extensions. Any such additional library functions would be extensions of the standard (whether such extensions are allowed by the standard is tangential to this question). – Dan Moulding Dec 10 '09 at 17:59
  • I should amend my previous comment: `-ansi` disables *conflicting* extensions, while `-pedantic` is supposed to disable all extensions. – Dan Moulding Dec 10 '09 at 18:01
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    Actually, I've looked at both C and C++ standards and although they allow additional library functions, such extensions cannot "alter the behavior of any well-formed program". This implies that they cannot reserve any additional identifiers (i.e. a well-formed program that defines its own round() function would break by the library extending the standard with its own round function). Any additional library functions must have names that are *already* reserved, such as `_Round()`. – Dan Moulding Dec 10 '09 at 18:17