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I've written code in Arduino that cycles through 3 LED color lights but it seems (code) bug prone so I'm trying to come up with a new way to write it. Due to the complexity I'm going to stick with the phesdo code of what I'm trying to do. Here it is:

If (red LED isn't max and green LED is 0 and blue LED is 0)
  {inc red LED; update dot matrix}
If (red LED is max and green LED isn't max and blue LED is 0)
  {inc green LED; update dot matrix}
If ((red LED is/has-been max but not 0 ) and green LED is max and blue LED is 0)
  {dec red; update dot matrix}
If (red LED is 0 and green LED is max and blue LED isn't max)
  {inc blue; update dot matrix}
If (red LED is 0 and (green LED is/has-been max but not 0) and blue LED is max)
  {dec green; update dot matrix}
If (red LED isn't Max and green LED is 0 and blue is Max )
  {inc red; update dot matrix}
If (red LED is Max and green LED is 0 and (blue LED is/has-been Max but not 0))
  {dec blue; update dot matrix}

Update LED Driver;

Note: For a visual it's a color wheel going red->orange->green->teal->blue->pink->repeat

The caveat is, all this is in a loop that is only run once before exiting to fetch other data. It then has to return to this loop and remember the color location it left off. Other wise it would be very easy to wrap all this in for loops and execute it linearly. As it is it has to inc or dec one color, understand it's color position if you will, update the led driver and then come back to inc or dec remembering where it left off. So does anyone have a better code methodologically, pseudo style, I can use besides this complex if statements style.

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3 に答える 3

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ええ、それは素晴らしいことではありません...私は次のようなことをします..

typedef struct{ //maybe stick this in a union, depending on what the compiler does to it.
    unsigned char r:2;
    unsigned char g:2;
    unsigned char b:2;
}color;
const int numOfColors = 3;
color colors[numOfColors] = {
    {2,0,0},//r
    {1,1,0},//o
    {0,2,0}//g
};

for(int i = 0 ; 1 ; i++)
{
    color c = colors[i%numOfColors];
    //set color
    //update
    //wait
}
于 2013-04-26T18:01:46.840 に答える
2

少なくとも私があなたのif声明を読んだとき、あなたには 7 つの異なる状態があり (おそらく 8 つあるはずですか?)、最後に到達すると、最初の状態に戻ります。

これは、小さなルックアップ テーブルを備えたカウンターとして実装するのは非常に簡単で、状態の LED が点灯する状態番号からマップされます。

于 2013-04-26T18:02:12.350 に答える