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Date: | Wed, 7 Jan 2004 20:13:29 -0500 |
Message-Id: | <200401080113.i081DT7q021456@envy.delorie.com> |
From: | DJ Delorie <dj AT delorie DOT com> |
To: | djgpp AT delorie DOT com |
In-reply-to: | <bti3vp$tn8$1@au-nws-0001.flow.com.au> (bdeck@lycos.co.uk) |
Subject: | Re: Does using '-std=c99' with gcc 3.3.x cause '__STRICT_ANSI__' to be defined? |
References: | <btarqo$kgr$1 AT au-nws-0001 DOT flow DOT com DOT au> <3FF9241B DOT 2060501 AT earthlink DOT net> <btcs0a$ti5$1 AT au-nws-0001 DOT flow DOT com DOT au> <btemt9$rf4$2 AT nets3 DOT rz DOT RWTH-Aachen DOT DE> <btfhok$int$1 AT au-nws-0001 DOT flow DOT com DOT au> <ubrpgb8t2 DOT fsf AT elta DOT co DOT il> <bti3vp$tn8$1 AT au-nws-0001 DOT flow DOT com DOT au> |
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> Understood. Hopefuly when things start looking complete, '-std=c99' won't > generate a definition that bypasses half the defines of the header files > anymore :-) Yes, it will. If you specifically request that gcc strictly comply with the c99 standard, all non-c99 definitions are removed from the headers. If you want the non-c99 stuff in the headers, use -std=gnu99 instead.
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