什么是CMake等价于“gcc -fvisibility = hidden”当控制共享库的导出符号？ [英] What is the CMake equivalent to "gcc -fvisibility=hidden" when controlling the exported symbol of a shared library?

问题描述

我在c ++中开发了跨平台软件。我知道，Linux .so默认导出所有符号，通过gcc -fvisibility = hidden我可以将所有导出的符号设置为隐藏，然后设置 __ attribute __（visibility（default））为我想导出的类和函数，所以我可以控制要导出的内容。

I developed cross platform software in c++. As I know, Linux .so exported all the symbols by default, well through "gcc -fvisibility=hidden" I can set all the exported symbols as hidden, then set __attribute__(visibility("default")) for the class and function I want to export, so I can control what I want to export.

我的问题是，使用CMake，我如何做gcc -fvisibility = hidden控件？

My question is, using CMake, how can I do the work as "gcc -fvisibility=hidden" control?

推荐答案

您可以向Cmake编译器添加一个标志：

You can add a flag to the Cmake compiler like that:

SET(CMAKE_C_FLAGS "${CMAKE_C_FLAGS} -fvisibility=hidden")

要确保这仅在Linux下完成，您可以使用此代码：

To make sure that this only done under Linux you can use this code:

IF(UNIX)
    IF(CMAKE_COMPILER_IS_GNUCC)
         SET(CMAKE_C_FLAGS "${CMAKE_C_FLAGS} -fvisibility=hidden")
    ENDIF(CMAKE_COMPILER_IS_GNUCC)
ENDIF(UNIX)

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