Error when using same library name as file

[erick166]

I'm generating a new VVC using UVVM's vvc_generator.py script, which generates the following files:

• test_bfm_pkg.vhd
• test_vvc.vhd
• vvc_cmd_pkg.vhd
• vvc_context.vhd
• vvc_methods_pkg.vhd

If the library to which I'm compiling the files is named test_vvc, which is the same as one of the files, I get the following error:
C:\uvvm\uvvm_vvc_framework\script\vvc_generator\output\test_vvc.vhd:70:35: error: no visible declaration for TEST_VVC.TD_CMD_QUEUE_PKG

[nickg]

Can you provide a self contained reproducer or the exact commands to run please?

[erick166]

I think the easiest is if you have UVVM to run the script yourself. You can find the script under uvvm_vvc_framework/script/vvc_generator/

1. From that location run: python3 vvc_generator.py
2. "Enter the VVC name:" test
3. "Generate with extended features:" n
4. "Set number of concurrent channels:" 1
5. "Multiple executors?:" n

It will generate the 5 files mentioned above, you don't need to compile vvc_context.
If you compile the files to a library named "test_vvc" you will get the error.