unknown identifier

Dear Sir

Something is wrong with my datafile and I don’t know how to solve the problem. The error information is : unknown identifier in data file: 15 12.01100 (read_data.cpp:1396). If I delete that line in the data file, the error information would turn to incorrect atom format, however, I think I wrote the right atom format.I tried several data files, non of them works, So I don’t know what to do. My datafile, in script and the error information have been attached, could you please take a look for me, I am so frustrated that I don’t know how to go on, I would really appreciate your help .

Best regards

Yihua Zhou

New folder.rar (47.9 KB)

You have two separate errors. First, the unknown identifier problem is because in your coordinate file (I opened coordinates.data), you specify that you have 14 atom types, but list 15 masses. You should have the same number of masses as you have atom types.

The second error, incorrect atom format, refers to the fact that you atom data in coordinates.data does not match the “full” style you specified next to “atom_style” in in.obtain_para_coeffs. You can learn more about what you need to include in your coordinate file for each atom_style here http://lammps.sandia.gov/doc/read_data.html

Hope that helps!

I’m afraid I’m not much help here. A seg fault can occur for a number of reasons and is an OS problem, not a code problem. My understanding is that it usually occurs when your code is trying to access memory that it cannot (I’m a chemist, not a computer scientist, so I’m not 100% on programming problems). It could be that your initial simulation set up has bad dynamics (too many atoms close together or overlapping maybe?). There may be other causes. It is hard to know without more information. What is the last step of the simulation or output before you get a segmentation fault? Also, when you reply, you should reply all so that this goes to the whole list. I’m sure there are many people on here that are more knowledgeable about programming and might be able to provide you more help than myself.