如何在可執行文件中使用函數時編譯.so。

Question

當我編譯libb.so它取決於liba.so。我可以運行

g++ -la. 

當libb.so使用的函數從可執行c，什麼是正確的命令？

g++ -lc 

拋出錯誤！

我想在android下編譯weston。 gl_render.so使用weston的函數。

Answer 1

我想編譯'libapi_simple.so'，它使用名爲'test'的可執行文件的函數int module_add（int，int）。 objs編譯和finnally cmd如下：

/code/phone/xmake/prebuilts/gcc/linux-x86/arm/arm-linux-gnueabi-4.9-glibc-2.20/bin/arm-linux- gnueabi-g ++ -shared -Wl，-soname，libapi_simple.so -fPIC -fPIE -o /code/phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/LINKED/libapi_simple.so/code /phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/libapi_simple_intermediates/api.o /code/phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/libapi_simple_intermediates /private.o \ -L/code/phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/LINKED \ -L/code/phone/out/platforms/phone.amlogic.eng .arm/target/obj/STATIC_LIBRARY \ -Wl， - no-undefined -Wl， - build-id -L/code/phone/out/platforms/phone.amlogic.eng。 arm/target/obj/SHARED_LIBRARY/LINKED -L/code/phone/out/platforms/phone.amlogic.eng.arm/target/rootfs/lib -L ​​/ code/phone/out/platforms/phone.amlogic.eng。 arm/target/rootfs/usr/lib \ -Wl，-rpath =/usr/lib \ -Wl，-rpath-link =/code/phone/out/platforms/phone.amlogic.eng.arm/target/OBJ/SHARED_LIBRARY /聯：/code/phone/out/platforms/phone.amlogic.eng.arm/target/rootfs/lib：/code/phone/out/platforms/phone.amlogic.eng.arm/target/rootfs/ usr/lib /code/phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/libapi_simple_intermediates/api.o：In function get_api_tag()': /code/phone/xmake/example/SHARED_LIBRARY_SIMPLE/api.cpp:9: undefined reference to module_add（int，int）' collect2：error：ld returned 1退出狀態 make：*** [/code/phone/out/platforms/phone.amlogic.eng.arm/target/obj/SHARED_LIBRARY/LINKED/libapi_simple.so]錯誤1

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