is the use of glfw not supported on arm linux?

I want to use the glfw library on the am5728 (kernel, run linux operating system) chip on the am5728 (mv7, run the linux operating system) and cross-compile on Ubuntu using the cmake-gui, The compiler is arm_linux_gnuabihf. Prompt: system is unknown to make, is the use of glfw not supported on arm linux? If I want to use, what should I do, thank you!

It looks like Raspberry Pi users have been able to get GLFW working on their arm linux systems by compiling on the target system rather than cross compiling, so I wonder if your issue is that your Ubuntu system or cmake does not have support for that particular cross compiling target system.

Can you compile on the target system?

Hi dougbinks,
Thanks for this quick answer.
I used cmake-gui to cross-compile glfw on ubuntu with an error. I did not compile directly on the target system.
The problem has been confirmed with chip manufacturer TI, who replied that I don’t support GLFW on am5728. I don‘’t know why. Is the implementation of GLFW related to the architecture of CPU? am5728 is armv7 architecture.

The error “system is unknown to cmake” is not a GLFW specific error, but one related to your cmake setup and command for cross compiling.

If you can compile directly on the system rather than cross compiling you might find this easier.

The chip manufacturer does not need to support GLFW directly, so long as they use provide the correct X11 and OpenGL development packages (which I think they do) you should be able to get this to work.

In summary, I do not think this is a GLFW specific issue but one related to cross compiling with cmake. I’m not an expert on cross compiling with cmake and suggest you first try compiling on the host system, and if that is not possible you might find searching for the error leads to a helpful answer.

I may be able to help more once you’ve tried the above, but will need more information your how you setup cmake, and what the full command line and output was. The toolchain file would also be useful.