Solution
Dependencies: mbed
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README.txt@1:facc2bd1d423, 2018-10-02 (annotated)
- Committer:
- noutram
- Date:
- Tue Oct 02 07:37:08 2018 +0000
- Revision:
- 1:facc2bd1d423
Added solution to tasks
Who changed what in which revision?
User | Revision | Line number | New contents of line |
---|---|---|---|
noutram | 1:facc2bd1d423 | 1 | Answer to questions |
noutram | 1:facc2bd1d423 | 2 | |
noutram | 1:facc2bd1d423 | 3 | Q1. 1 second |
noutram | 1:facc2bd1d423 | 4 | Q2. XOR |
noutram | 1:facc2bd1d423 | 5 | Q3. D3 |
noutram | 1:facc2bd1d423 | 6 | |
noutram | 1:facc2bd1d423 | 7 | Q4. This project is one possible solution |
noutram | 1:facc2bd1d423 | 8 | |
noutram | 1:facc2bd1d423 | 9 | The key line is this |
noutram | 1:facc2bd1d423 | 10 | |
noutram | 1:facc2bd1d423 | 11 | unsigned int binaryInput = SW1 + (SW2 << 1); |
noutram | 1:facc2bd1d423 | 12 | |
noutram | 1:facc2bd1d423 | 13 | where SW1 and SW2 are the switches for the least and most significant bits respectively. |
noutram | 1:facc2bd1d423 | 14 | |
noutram | 1:facc2bd1d423 | 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). |