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我正在使用带有两个压电元件的独立ATmega328P来生成一些音乐。

我已经用音符的频率定义了一些常数。然后我定义了一个结构，其中包含第一个和第二个压电的注释以及注释的长度。然后我制作了更多这些结构的数组来描述每首歌曲。

问题是这样我很快就会耗尽内存。我试图将结构数组存储在 PROGMEM 中，以避免这个问题。我尝试使用一个名为 PROGMEM_readAnything 的小型库、memcpy_P() 或 pgm_read_word() 和 pgm_read_byte() 函数，但在所有情况下我都会遇到同样的问题。

当我遍历 NOTES 数组时，它会跳过一些元素，而正确读取和播放其他元素。它总是跳过相同的元素，而不仅仅是随机元素。

我什至尝试更换微控制器，认为芯片的某些部分可能已被某些东西损坏，但上传相同的草图我得到了相同的结果，因此微控制器可能完好无损。

这是代码：

#include <Tone.h>
#include <avr/pgmspace.h>

// Define the notes frequency
#define G2 98
#define Gs2 104
#define Ab2 104
#define A2 110
#define As2 116

//... and so on with many other music notes ...

#define Fs7 2960
#define Gb7 2960
#define G7 3136

//Rest
#define R 0

typedef struct {
    int n1;
    int n2;
    byte units;
} NOTES;

Tone buzzer1;
Tone buzzer2;

int myTempo = 100;

// Walkyrie
const NOTES walkyrie[] PROGMEM = {

                          {Fs3, Fs4, 2},
                          {B3,  B4,  3},
                          {Fs3, Fs4, 1},
                          {B3,  B4,  2},
                          {D4,  D5,  6},
                          {B3,  B4,  6},
                          {D4,  D5,  3},
                          {B3,  B4,  1},
                          {D4,  D5,  2},
                          {Fs4, Fs5, 6},
                          {D4,  D5,  6},
                          {Fs4, Fs5, 3},
                          {D4,  D5,  1},
                          {Fs4, Fs5, 2},
                          {A4,  A5,  6},
                          {A3,  A4,  6},
                          {D4,  D5,  3},
                          {A3,  A4,  1},
                          {D4,  D5,  2},
                          {Fs4, Fs5, 6},
                          {R,    0,  4},
                          {A3,  A4,  2},
                          {D4,  D5,  3},
                          {A3,  A4,  1},
                          {D4,  D5,  2},
                          {Fs4, Fs5, 6},
                          {D4,  D5,  6},
                          {Fs4, Fs5, 3},
                          {D4,  D5,  1},
                          {Fs4, Fs5, 2},
                          {A4,  A5,  6},
                          {Fs4, Fs5, 6},
                          {A4,  A5,  3},
                          {Fs4, Fs5, 1},
                          {A4,  A5,  2},
                          {Cs5, Cs6, 6},
                          {Cs4, Cs5, 6},
                          {Fs4, Fs5, 3},
                          {Cs4, Cs5, 1},
                          {Fs4, Fs5, 2},
                          {As4, As5, 6}
                        };

void playSong()
{
    // We store the frequency of the second piezo in this variable
    int secondFreq = 0;
    Serial.println(sizeof(walkyrie)/sizeof(walkyrie[0]));
    // Walk through the array of music
    for(int i = 0; i < sizeof(walkyrie)/sizeof(walkyrie[0]); i++)
    {
        int n1;
        int n2;
        byte units;
        // Only play if it is not a rest
        if (walkyrie[i].n1 > 0)
        {
            n1 = pgm_read_word(&(walkyrie[i].n1));
            n2 = pgm_read_word(&(walkyrie[i].n2));
            units = pgm_read_byte(&(walkyrie[i].units));

            Serial.print("Row ");
            Serial.print(i);
            Serial.print(": Frequency 1: ");
            Serial.print(n1);
            Serial.print(" Frequency 2: ");
            Serial.print(n2);
            Serial.print(" Units: ");
            Serial.println(units);

            // Play the note of the first piezo
            buzzer1.play(n1, (units*myTempo));
            // If the frequency of the second piezo is 0, we play the same note
            // as the first, else the note set for the second one
            if (n2 == 0)
            {
                secondFreq = n1;
            }
            else {
                secondFreq = n2;
            }

            buzzer2.play(secondFreq, (units*myTempo));
        }

        // Then we wait for the note to end plus a little, between two notes
        delay((units*myTempo) + 10);
    }
}


void setup() {
    Serial.begin(9600);
    buzzer1.begin(11);
    buzzer2.begin(12);
}

void loop()
{
     playSong();
}

我添加了一些行以在串行监视器中查看会发生什么。它读取正确的长度...

串行监视器的输出如下：

41                                        (correct length)
Row 1: Freq1: 247 Freq2: 499 Units: 3     (row 0 - the first note is already missing)
Row 2: Freq1: 185 Freq2: 370 Units: 1
Row 3: Freq1: 247 Freq2: 499 Units: 2     (row 4 missing)
Row 5: Freq1: 247 Freq2: 499 Units: 6     (row 6-7 missing)
Row 8: Freq1: 294 Freq2: 587 Units: 2
Row 9: Freq1: 370 Freq2: 740 Units: 6
Row 10: Freq1: 294 Freq2: 587 Units: 6
Row 11: Freq1: 370 Freq2: 740 Units: 3
Row 12: Freq1: 294 Freq2: 587 Units: 1
Row 13: Freq1: 370 Freq2: 740 Units: 2
Row 14: Freq1: 440 Freq2: 880 Units: 6
Row 15: Freq1: 220 Freq2: 440 Units: 6    (row 16-17 missing)
Row 18: Freq1: 294 Freq2: 587 Units: 2
Row 19: Freq1: 370 Freq2: 740 Units: 6
Row 20: Freq1: 0 Freq2: 0 Units: 4
Row 21: Freq1: 220 Freq2: 440 Units: 2
Row 22: Freq1: 294 Freq2: 587 Units: 3
Row 23: Freq1: 220 Freq2: 440 Units: 1
Row 24: Freq1: 294 Freq2: 587 Units: 2
Row 25: Freq1: 370 Freq2: 740 Units: 6
Row 26: Freq1: 294 Freq2: 587 Units: 6
Row 27: Freq1: 370 Freq2: 740 Units: 3
Row 28: Freq1: 294 Freq2: 587 Units: 1
Row 29: Freq1: 370 Freq2: 740 Units: 2
Row 30: Freq1: 440 Freq2: 880 Units: 6
Row 31: Freq1: 370 Freq2: 740 Units: 6
Row 32: Freq1: 440 Freq2: 880 Units: 3
Row 33: Freq1: 370 Freq2: 740 Units: 1
Row 34: Freq1: 440 Freq2: 880 Units: 2
Row 35: Freq1: 554 Freq2: 1109 Units: 6
Row 36: Freq1: 277 Freq2: 554 Units: 6
Row 37: Freq1: 370 Freq2: 740 Units: 3
Row 38: Freq1: 277 Freq2: 554 Units: 1
Row 39: Freq1: 370 Freq2: 740 Units: 2
Row 40: Freq1: 466 Freq2: 932 Units: 6

为什么会这样？或者有没有更好、更有效的方法来解决这个问题？

4

1 回答 1

3

在这一行中，您检查了数据，但您没有执行“pgm_read_word()”来实际从闪存中获取数据：

if(walkyrie[i].n1 > 0)

如果意外地得到一个非零值，那么您正确地从闪存中读取了这些值，否则，您将跳过该行。

进一步的证据：

Row 20: Frq1: 0 Frq2: 0 Units: 4

在这里，n1为零，但该测试应该跳过该行。

此外，“休息”的逻辑有点不对劲。现在，您不会在其余时间读取单位，因此它使用之前的值（来自演奏的音符）。

我想我会先得到所有三个值，然后检查它们。

我还将频率编码为一个字节，并使用查找表将“键号”转换为频率（如 MIDI 键号）。这样，您的结构数组会小一些。也许也打开 __packed__ （随便）属性，以避免条目之间的填充 - 如果节省闪存空间很重要（那么你可以在那里获得更多歌曲！）

听起来很有趣！祝你好运！

于 2016-08-11T21:03:46.403 回答