Question

我有一个程序，我想创建一个随机生成的音乐栏（4个节拍到一个小节，使用C大调音阶）。但是，我在理解数学方面遇到了麻烦，并且在循环中不断溢出，为条形码创建了4个以上的音符。

我正在使用由我的导师创建的aServe，但基本上会向一个振荡器打开一个流，该播放器播放我所评论过的论据。

/* Program for randomly written bar of 4/4 in C Major */

#include "aservelibs/aservelib.h"
#include <stdio.h>
#include <stdlib.h>

//macros
#define SEMIBREVE       (1.0)
#define MINIM           (1.0/2)
#define CROTCHET        (1.0/4)
#define QUAVER          (1.0/8)
#define SEMIQUAVER      (1.0/16)
#define DEMISEMIQUAVER  (1.0/32)

#define C   (261.63)
#define D   (293.66)
#define E   (329.63)
#define F   (349.23)
#define G   (391.99)
#define A   (440.00)
#define B   (493.88)

int millisec(int bpm, double note) {
    return (int)(
                 60      /* seconds */
                 * 1000  /* milliseconds per second */
                 * 4     /* crotchets per semibreve */
                 * note
                 / bpm
                 );
}

int main()
{
    int bpm = 120; //BPM Value
    double Length[] = {SEMIBREVE, MINIM, CROTCHET, QUAVER, SEMIQUAVER, DEMISEMIQUAVER}; //Array of Note Lengths
    double Pitch[] = {C, D, E,F, G, A, B}; //Array of CMajor Scale Freq

    int randLength = (rand() % 6); //random positions for note length
    int randPitch = ( rand() % 7); //random positions for note pitch
    double barTotal = 0; //amount of bar currently completed

    do {
        if(barTotal < 1) //if bar total is smaller than 1
        {
            barTotal = Length[randLength] + barTotal; //add note to total

            aserveOscillator(0, Pitch[randPitch], 1, 2); //Starts stream to oscialltor
                //aserveOscillator(Index,Frequency,Amplitude,WaveType);
            aserveSleep(millisec(bpm, Length[randLength])); //play the notes for the length of time specified in milliseconds

            randLength = (rand() % 6); //prepare next random note
            randPitch = (rand() % 7); //prepare next random pitch

            //Output
            printf("Note: ");
            printf("%lf", Pitch[randPitch]);
            printf("\n For: ");
            printf("%lf", Length[millisec(bpm,randLength)]);
            printf("\n With Bar Total: ");
            printf("%lf", barTotal);
            printf("\n\n");
        }
        else
        {
            if(barTotal != 1) //if bar total is bigger than 4
            {
                randLength = (rand() % 6); //try another number
            }
        }

    } while (barTotal != 1); //will stop once reaches 4

    return 0;
}

Answer 1

考虑以不同的方式思考问题。将条形图视为“n”个插槽，其中n是您具有的最细粒度的音符类型。因此，在您的情况下，条形图是一组32个插槽。不是将数字表示为分数，而是使用整数类型来显示每个“槽”的数量。因此DEMISEMIQUAVER需要1个插槽，可以表示为int而不是（1.0 / 32.0），这会引入一些可能很难看的问题。

一旦你这样做，解决方案就更简单了：

1）当前栏中剩余多少个插槽？ 2）从小于剩余插槽的音符池中选择随机音符 3）重新计算添加新笔记后剩余的空间 4）如果剩余房间为零，请前往下一个栏。

以下是您的代码，适用于这种新方法。没有经过充分测试，但它应该避免到目前为止讨论的大部分（如果不是全部）陷阱。

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

//macros
#define SEMIBREVE       (32)
#define MINIM           (16)
#define CROTCHET        (8)
#define QUAVER          (4)
#define SEMIQUAVER      (2)
#define DEMISEMIQUAVER  (1)

#define C   (261.63)
#define D   (293.66)
#define E   (329.63)
#define F   (349.23)
#define G   (391.99)
#define A   (440.00)
#define B   (493.88)

int GetMaxIndex(int remainingLength)
{
    // Returns the largest upper bound of the Length array that
    // should be considered based on how much room remains in 
    // the current bar.
    int result;
    if(remainingLength == 32) result = 5;
    if(remainingLength < 32) result = 4;
    if(remainingLength < 16) result = 3;
    if(remainingLength < 8) result = 2;
    if(remainingLength < 4) result = 1;
    if(remainingLength < 2) result = 0;
    return result;

}

int main()
{
    double Pitch[] = {C, D, E,F, G, A, B}; //Array of CMajor Scale Freq

    int bpm = 120; //BPM Value
    int Length[] = {DEMISEMIQUAVER, SEMIQUAVER, QUAVER, CROTCHET, MINIM, SEMIBREVE}; //Array of Note Lengths
    char* Labels[] = {"DEMISEMIQUAVER (Thirty Second)", "SEMIQUAVER (Sixteenth)", "QUAVER (Eighth)", "CROTCHET (Quarter)", "MINIM (Half)", "SEMIBREVE (Whole)"}; 
    int remainingThisBar;
    int barsToGenerate = 4;

    int randLength = (rand() % 6); //random positions for note length
    int randPitch; //random positions for note pitch
    int maxIndex;
    int randIndex;

    srand(time(NULL));

    for(int barNumber = 0; barNumber < barsToGenerate; barNumber++)
    {
        printf("Beginning bar: %i\n", barNumber);
        remainingThisBar = 32;

        while(remainingThisBar > 0)
        {
            maxIndex = GetMaxIndex(remainingThisBar);  // What is the biggest note index we still have room for?
            randIndex = maxIndex == 0 ? 0 : (rand() % maxIndex); // Get a random note between 0 and maxIndex

            randPitch = ( rand() % 7);      // Random positions for note pitch
            randLength = Length[randIndex]; // Length in 32nds
            remainingThisBar -= randLength; 

            // Output
            printf("\tNote: %s @ %f\n", Labels[randIndex], Pitch[randPitch]);
            printf("\t32nds remaining in bar: %i\n", remainingThisBar);
            printf("\n");

            /* TODO - Output note via aServe*/
        }
    }
}

如何解决生成音乐的无限循环？

1 个答案: