University of Plymouth - Stages 1, 2 and 3
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Task352Solution
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Fork of
Task352Solution
by 
Nicholas Outram
README.txt@2:82201f763676, 2018-10-02 (annotated)
- Committer:
- noutram
- Date:
- Tue Oct 02 07:33:59 2018 +0000
- Revision:
- 2:82201f763676
Added solutions to questions in notes
Who changed what in which revision?
| User | Revision | Line number | New contents of line |
|---|---|---|---|
| noutram | 2:82201f763676 | 1 | Answer to questions |
| noutram | 2:82201f763676 | 2 | |
| noutram | 2:82201f763676 | 3 | Q1. 1 second |
| noutram | 2:82201f763676 | 4 | Q2. XOR |
| noutram | 2:82201f763676 | 5 | Q3. D3 |
| noutram | 2:82201f763676 | 6 | |
| noutram | 2:82201f763676 | 7 | Q4. This project is one possible solution |
| noutram | 2:82201f763676 | 8 | |
| noutram | 2:82201f763676 | 9 | The key line is this |
| noutram | 2:82201f763676 | 10 | |
| noutram | 2:82201f763676 | 11 | unsigned int binaryInput = SW1 + (SW2 << 1); |
| noutram | 2:82201f763676 | 12 | |
| noutram | 2:82201f763676 | 13 | where SW1 and SW2 are the switches for the least and most significant bits respectively. |
| noutram | 2:82201f763676 | 14 | |
| noutram | 2:82201f763676 | 15 | Note the use of << to multiply by 2. A left shift is a cheap operation in computing terms (does not require many CPU cycles). |