GCC创建一个共享对象而不是可执行二进制文件 [英] GCC creates a shared object instead of an executable binary

问题描述

我有一个我正在建造的图书馆。当我运行以下任何一个对象时，所有对象都会依次编译和链接：
ar rcs lib / libryftts.a $ ^

gcc -shared $ ^ -o lib / libryftts.so

在我的Makefile中。我也能够成功地将它们安装到 / usr / local / lib
当我使用nm测试文件时，所有函数都在那里。
我的问题是，当我运行 gcc testing / test.c -lryftts -o test&&文件./test 或 gcc testing / test.c lib / libryftts.a -o test&&文件./test
它说：

test：ELF 64位LSB共享对象 code>而不是 test：ELF 64位LSB可执行文件，如我所料。我做错了什么？

解决方案

我做错了什么？

无。

听起来像您的GCC被配置为构建 - 馅饼默认情况下为二进制文件。这些二进制文件实际上是共享库（类型为 ET_DYN ），除非它们像普通可执行文件那样运行。

所以你应该运行你的二进制文件，（如果它有效的话）不用担心它。

或者你可以将你的二进制文件与 gcc -no-pie ... 并且应该产生一个非< PIE 类型的可执行文件 ET_EXEC ，其中文件会说 ELF 64位LSB可执行文件。

I have a library I am building. All of my objects compile and link successively when I run either one of: ar rcs lib/libryftts.a $^

gcc -shared $^ -o lib/libryftts.so

in my Makefile. I also am able to successfully install them into /usr/local/lib When I test the file with nm, all the functions are there. My problem is that when I run gcc testing/test.c -lryftts -o test && file ./test or gcc testing/test.c lib/libryftts.a -o test && file ./test it says:

test: ELF 64-bit LSB shared object instead of test: ELF 64-bit LSB executable as I would expect. What am I doing wrong?

解决方案

What am I doing wrong?

Nothing.

It sounds like your GCC is configured to build -pie binaries by default. These binaries really are shared libraries (of type ET_DYN), except they run just like a normal executable would.

So your should just run your binary, and (if it works) not worry about it.

Or you could link your binary with gcc -no-pie ... and that should produce a non-PIE executable of type ET_EXEC, for which file will say ELF 64-bit LSB executable.

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