svn commit: r362304 - head/x11-toolkits/pango

[Tijl Coosemans]

On Mon, 21 Jul 2014 15:00:52 +0200 John Marino wrote:
> On 7/21/2014 14:43, Tijl Coosemans wrote:
>> On Mon, 21 Jul 2014 14:15:55 +0200 John Marino wrote:
>>> On 7/21/2014 13:26, Tijl Coosemans wrote:
>>>> On Mon, 21 Jul 2014 12:55:23 +0200 John Marino wrote:
>>>>> Everything that uses a pango function that has a libm symbol must also
>>>>> link with libm.
>>>>
>>>> This is a completely false statement.  If X links to Y and Y uses Z
>>>> symbols, you do not have to link X with Z.  Y links with Z and that is
>>>> enough.  Otherwise X would have to link with its entire dependency
>>>> tree.
>>>
>>> If the linker doesn't follow Y's link to Z, how is it supposed to
>>> resolve Z references?
>> 
>> If X doesn't contain Z references the linker doesn't have to resolve
>> Z references.
>> 
>> If X does contain Z references then explicit linking requires that you
>> explicitly link X with -lZ and that you cannot rely on -lY to imply -lZ.
> 
> This seems to be the heart of our disagreement.
> I am saying X can pull in a function of Y that contains a symbol of Z.
> In that case, there's no reference of Z in X, but when linking X it
> still needs -lZ.

Let's work out an example:

-- X.c --
void funcY( void );
int main( void ) {
	funcY();
	return( 0 );
}
---------

-- Y.c --
void funcZ( void );
void funcY( void ) {
	funcZ();
}
---------

-- Z.c --
#include <stdio.h>
void funcZ( void ) {
	( void ) puts( "Hello world" );
}
---------

Create libZ.so:
% cc -shared -o libZ.so Z.c

Create libY.so linking it with libZ:
% cc -shared -o libY.so Y.c -Wl,-rpath,. -L. -lZ

Create X linking it with libY, but not libZ:
% cc -o X X.c -Wl,-rpath,. -L. -lY

Run ./X:
% ./X
Hello world


Now, what do you get?