C: WAV 文件未在 "rb" 中打开

Question

我在尝试打开 WAV 文件进行阅读时遇到困难。当我编译运行我的代码时，我没有收到任何错误。我正在添加我正在处理的代码（假设已经调用了必要的库）。该程序应该显示 WAV 文件的内容，尽管我输入了有效的文件名和扩展名，但语句 "Invalid Filename. Try Again." 仍在屏幕上打印。我尝试的另一种方法是输入文件的目录而不仅仅是名称，当我这样做时，我的程序结束并且没有显示任何内容。任何指导都会有所帮助，在此先感谢您！

main(){
FILE *fin;
printf("\nEnter filename of WAV file: \n");
char filename[256];
scanf("%s",&filename);
fin = fopen(filename,"rb"); // opens in rb

if(!fin) // if file doesn't exist
{
printf("Invalid filename. Try again.\n");
}
else // if fin opens succesfully
{
    printf("\nFile opened succesfully\n");

    char *header;
    header = (char *)malloc(44);
    if(header == NULL)
    {
        printf("Error in allocating memory.");
        return 0;
    }


    fread(header,1,44,fin);

    char *chunkid;
    unsigned int *chunksize;
    char *format;
    char *subchunk1id;
    unsigned int *subchunk1size;
    unsigned short int *audioformat;
    unsigned short int *numchannels;
    unsigned int *samplerate;
    unsigned int *byterate;
    unsigned short int *blockalign;
    unsigned short int *bitspersample;
    char *subchunk2id;
    unsigned int *subchunk2size;
    unsigned int *data;

    chunkid = header;
    chunksize = (unsigned int *)(header + 4);
    format = header + 8;
    subchunk1id = header + 12;
    subchunk1size = (unsigned int *)(header + 16);
    audioformat = (unsigned short int *)(header + 18);
    numchannels = (unsigned short int*)(header + 20);
    samplerate = (unsigned int*)(header + 24);
    byterate = (unsigned int*)(header + 28);
    blockalign = (unsigned short int*)(header + 30);
    bitspersample = (unsigned short int*)(header + 32);
    subchunk2id = header + 36;
    subchunk2size = (unsigned int*)(header + 40);
    data = (unsigned int*)(header + 44);

    printf("\n%c%c%c%c",*(header),*(header+1),*(header+2),*(header+3));
    printf("\n%d",*chunksize);
    printf("\n%c%c%c%c",*(header + 8),*(header + 9), *(header + 10), *(header + 11));
    printf("\n%c%c%c%c",*(header + 12),*(header + 13), *(header + 14), *(header + 15));
    printf("\n%d",*subchunk1size);
    printf("\n%d",*audioformat);
    printf("\n%d",*numchannels);
    printf("\n%d",*samplerate);
    printf("\n%d",*byterate);
    printf("\n%d",*blockalign);
    printf("\n%d",*bitspersample);
    printf("\n%c%c%c%c",*(header + 36),*(header + 37),*(header + 38), *(header + 39));
    printf("\n%d",*subchunk2size);
    printf("\n%d",*data);


    fclose(fin);

} // end of else
} // end of main

Answer 1

尝试使用

if(!fin)
{
  perror("Error:");
}

它会打印出正确的错误消息，以便您轻松诊断无法打开文件的原因。可能是您的程序找不到您指定的文件，或者您可能没有权限打开它。 Here 列出了您在打开文件时可能遇到的错误。

我能看到的另一个可能的问题是您正在使用 scanf 从 stdin 读取到您的 char 数组。如果你指定的路径包含white-space怎么办？后面跟有 White-space 的每个单词都将被视为不同的字符串。你只会得到第一个最长的不包含 white-space 的单词。所以，你的程序没有读取整个路径。你可以试试 scanf("%[^\n]", filename) 将读取直到遇到 '\n'。谨防！这使您面临缓冲区溢出的可能性，如果您将一个有效的字符串集传递给长度大于 256 的程序，您将超出 char 数组的边界进行写入，这并不好。所以尝试 scanf("%255[^\n]", filename) 或尝试使用提供缓冲区溢出保护的 fgets() doc SO。

Answer 2

在编译代码时使用警告会暴露你的错误。

改进建议，将 header 描述为如下结构：

typedef struct  
{
    char    riff_tag[4];
    int32_t riff_length;
    char    wave_tag[4];
    char    fmt_tag[4];
    int32_t fmt_length;
    int16_t audio_format;
    int16_t num_channels;
    int32_t sample_rate;
    int32_t byte_rate;
    int16_t block_align;
    int16_t bits_per_sample;
    char    data_tag[4];
    int32_t data_length;
}waveFileHeader, *waveFileHeaderptr ;

并像这样阅读：

waveFileHeaderptr header =malloc ( sizeof (waveFileHeader));
fread (header , sizeof (waveFileHeader), 1 , fin);

将使以下代码和处理不那么混乱

C: WAV 文件未在 "rb" 中打开

C: WAV file not being opened in "rb"

c

audio

binary

file

wav