Problems with RTSC packages and legacy binary objects [message #1670] |
Thu, 12 March 2009 21:10  |
Eclipse User |
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I have a RTSC package (say ti.mas.pkgA) that advertises some legacy
libraries (say lib1.a and lib2.a) in its getLibs() function in package.xs.
The problem is that there are circular references between lib1.a and
lib2.a. Another package ti.mas.pkgB is an application package needs to
link these libraries to build the application.
My getLibs looks like this:
function getLibs(prog)
{
var lib="";
lib = lib1.a; lib2.a ;
return (lib);
}
When pkgB is built, I get linker errors due to the circular references
between lib1 and lib2. To resolve this, I really want them to be placed in
a linker group like this:
-Wl,--start-group lib1.a lib2.a Wl,--end-group
I cannot figure out how to do this in RTSC. Any ideas?
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Re: Problems with RTSC packages and legacy binary objects [message #1702 is a reply to message #1670] |
Mon, 16 March 2009 11:23  |
Eclipse User |
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Chander Raja wrote:
> I have a RTSC package (say ti.mas.pkgA) that advertises some legacy
> libraries (say lib1.a and lib2.a) in its getLibs() function in
> package.xs. The problem is that there are circular references between
> lib1.a and lib2.a. Another package ti.mas.pkgB is an application package
> needs to link these libraries to build the application.
>
> My getLibs looks like this:
> function getLibs(prog)
> {
> var lib="";
>
> lib = �lib1.a; lib2.a� ;
>
> return (lib);
> }
>
>
> When pkgB is built, I get linker errors due to the circular references
> between lib1 and lib2. To resolve this, I really want them to be placed
> in a linker group like this:
>
>
> -Wl,--start-group lib1.a lib2.a �Wl,--end-group
>
>
> I cannot figure out how to do this in RTSC. Any ideas?
>
Can you create a third library which places the object of both lib1 and
lib2 into a single library, say lib3? Legacy users can directly link
with lib1 and lib2 while getLibs returns lib3.
This should have the effect of the --startgroup --endgroup without
relying on a TI-specific linker feature.
>
>
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