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If you want an embedded programming, OS, or firmware job, your best bet is with a hardware/semiconductor company. No, you don't need a masters or doctorate --
by pdq 11y ago
If you want an embedded programming, OS, or firmware job, your best bet is with a hardware/semiconductor company. No, you don't need a masters or doctorate -- a CS degree is just fine. I highly recommend taking a low level job; you learn an amazing amount that is abstracted away at the high-level web/database areas.
Have a look at the big hardware vendors: Qualcomm, Broadcom, Freescale (NXP), Nvidia, Cirrus Logic, Xilinx, Apple, Intel, AMD, etc. Note that you'd probably need to be willing to move to a more hardware-oriented area, including California, Silicon Valley, Austin, Boston, or Portland.
- vonmoltke 11y agoYou then have to be willing to deal with lots of silence. I can't seem to get an interview with any of those companies to save my life. They seem to be much pickier about whom they interview than web, mobile, and enterprise shops.
- djent 11y agoNvidia's careers site is the best yet - all the jobs for new college graduates require a 3.5 GPA!
- ajdlinux 11y agoWhile you're probably going to have more luck in the more hardware-oriented places, that's not to say there are no OS/firmware jobs elsewhere - I got a job straight out of my CS degree working on Linux at IBM in Canberra, of all places. I didn't do any work on the Linux kernel (outside of doing a single OS class) while I was studying, but certainly if you're able to contribute to a low-level open source project it's definitely not a bad thing.
- yitchelle 11y agoIt would be wise to look at the big players for specific industries. As an example for automotive, look at the tier 1 suppliers such as Bosch, Delphi, Continential or Johnson Controls, or even the vehicle manufacturers themselves who does software work, like Volvo Cars, Jaguar Land Rover, General Motors, Renault Trucks Go through industry you are interested in, do some googling and a list of potential companies can be found.