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Date: | Fri, 09 Feb 2001 12:30:05 +0200 |
From: | "Eli Zaretskii" <eliz AT is DOT elta DOT co DOT il> |
Sender: | halo1 AT zahav DOT net DOT il |
To: | rpolzer AT web DOT de (Rudolf Polzer) |
Message-Id: | <9791-Fri09Feb2001123005+0200-eliz@is.elta.co.il> |
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CC: | djgpp AT delorie DOT com |
In-reply-to: | <slrn987a3t.qp.rpolzer@rebounce.rpolzer-lx> |
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> <slrn987a3t DOT qp DOT rpolzer AT rebounce DOT rpolzer-lx> |
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> From: rpolzer AT web DOT de (Rudolf Polzer) > Newsgroups: comp.os.msdos.djgpp > Date: Fri, 9 Feb 2001 09:21:17 +0100 > > I forgot to ask, > is "gcc -O3 -ggdb3 -o X.EXE" and then "strip -s X.EXE" equivalent to > leaving out -ggdb3? No. "strip foo.exe" (note: no -s switch to strip) is actually _more_ than just leaving out -ggdb3. -ggdb3 adds line number and other source-related information, but a minimal set of information about symbols is put into the binary even without -g. `strip' removes _all_ of the symbolic info, leaving just the bare machine code. For example, without -ggdb3 you should still be able to run SYMIFY and get at least the functions' names from the addresses. After `strip', this is impossible.
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