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Re: How can I get -std=c99 without definine __STRICT_ANSI__?

From: "Richard Li" <richardpku at gmail dot com>
To: "John Love-Jensen" <eljay at adobe dot com>
Cc: "Rick Mann" <rmann at latencyzero dot com>, "MSX to GCC" <gcc-help at gcc dot gnu dot org>
Date: Sat, 22 Sep 2007 10:17:00 +0800
Subject: Re: How can I get -std=c99 without definine __STRICT_ANSI__?
References: <BA5D6032-0791-4916-AF6F-548465B07D3E@latencyzero.com> <C318322B.253E9%eljay@adobe.com>

Hi,

I think the current default is still gnu89, since GCC has not yet
fully implemented C99. Some C99 features (like variable declaration
after some statements) are supported by default as part of the GNU
extension to C89, but others (like variable declaration in a "for"
statement) are disable unless you specify "-std=c99" or "-std=gnu99"
explicitly.

On 9/21/07, John Love-Jensen <eljay@adobe.com> wrote:
> If you want C99 plus GNU extensions, use: -std:gnu99 (which is the default
> for C code, if not specified)

References:
- How can I get -std=c99 without definine __STRICT_ANSI__?
  - From: Rick Mann
- Re: How can I get -std=c99 without definine __STRICT_ANSI__?
  - From: John Love-Jensen

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