如何实现CS50的pset3频率和理解notes.c
如何实现CS50的pset3频率和理解notes.c
提问于 2018-06-04 00:38:33
我对CS50的音乐有几个问题。
1)在notes.c中有一行for (int i = 0, n = sizeof(NOTES) / sizeof(string); i < n; i++)。什么是sizeof(string)？我可以理解sizeof(注释)，这里的注释是一个字符串数组。但是在sezeof(string)？中string意味着什么呢？我拿不到。
2)下面是我对“频率”的实现：
int frequency(string note)
{
char N[2];
int octave;
//parsing the string into a note...
strncpy(N, note, (strlen(note) - 1));
//...and its octave
octave = note[strlen(note) - 1] - '0';
//add semitones
int semitone;
if (strcmp(N, "C") == 0)
semitone = 1;
if (strcmp(N, "C#") == 0 || strcmp(N, "Db") == 0)
semitone = 2;
if (strcmp(N, "D") == 0)
semitone = 3;
if (strcmp(N, "D#") == 0 || strcmp(N, "Eb") == 0)
semitone = 4;
if (strcmp(N, "E") == 0)
semitone = 5;
if (strcmp(N, "F") == 0)
semitone = 6;
if (strcmp(N, "F#") == 0 || strcmp(N, "Gb") == 0)
semitone = 7;
if (strcmp(N, "G") == 0)
semitone = 8;
if (strcmp(N, "G#") == 0 || strcmp(N, "Ab") == 0)
semitone = 9;
if (strcmp(N, "A") == 0)
semitone = 10;
if (strcmp(N, "A#") == 0 || strcmp(N, "Bb") == 0)
semitone = 11;
if (strcmp(N, "B") == 0)
semitone = 12;
//calculate freq: semitones
float freq = 440 * (powf(2, (semitone -10) / (float)12));
//calculate freq: multiply by num of octaves
return round(freq * (powf(2, octave - 4)));
}./note之后的输出为：
C4: 262
C#4: 922746880
D4: 294
helpers.c:55:12: runtime error: value 5.85908e+09 is outside the range of representable values of type 'int'
D#4: -2147483648
E4: 330
F4: 349
F#4: -2147483648
G4: 392
G#4: -2147483648
A4: 440
A#4: -2147483648
B4: 494代码本身可以工作，但在notes.c中不能工作。
请帮我了解一下到底出了什么问题。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50668632
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