如何强制 gcc 链接未使用的静态库
我有一个程序和一个静态库:
I have a program and a static library:
// main.cpp
int main() {}
// mylib.cpp
#include <iostream>
struct S {
S() { std::cout << "Hello World
";}
};
S s;
我想将静态库(libmylib.a)链接到程序对象(main.o)，虽然后者不使用前者的任何符号直接.
I want to link the static library (libmylib.a) to the program object (main.o), although the latter does not use any symbol of the former directly.
以下命令似乎不适用于 g++ 4.7.它们将在没有任何错误或警告的情况下运行，但显然 libmylib.a 不会被链接:
The following commands do not seem to the job with g++ 4.7. They will run without any errors or warnings, but apparently libmylib.a will not be linked:
g++ -o program main.o -Wl,--no-as-needed /path/to/libmylib.a
或
g++ -o program main.o -L/path/to/ -Wl,--no-as-needed -lmylib
你有更好的想法吗?
推荐答案
使用 --whole-archive 链接器选项.
在命令行中之后的库不会丢弃未引用的符号.您可以通过在这些库之后添加 --no-whole-archive 来恢复正常的链接行为.
Libraries that come after it in the command line will not have unreferenced symbols discarded. You can resume normal linking behaviour by adding --no-whole-archive after these libraries.
在您的示例中，命令将是:
In your example, the command will be:
g++ -o program main.o -Wl,--whole-archive /path/to/libmylib.a
一般来说，它将是:
g++ -o program main.o
-Wl,--whole-archive -lmylib
-Wl,--no-whole-archive -llib1 -llib2
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