我正在处理一个wav文件的幅度和缩放一些十进制的因素。我试图围绕如何以有效的内存方式读取和重写文件,同时试图解决语言的细微差别(我是C的新手)。该文件可以是8位或16位格式。我认为这样做的方法是首先将header data读入一些预定义的结构中,然后在循环中处理实际数据,在该循环中我将读取一块数据到缓冲区中,执行所需的任何操作,以及然后将其写入输出。处理音频wav文件与C
#include <stdio.h>
#include <stdlib.h>
typedef struct header
{
char chunk_id[4];
int chunk_size;
char format[4];
char subchunk1_id[4];
int subchunk1_size;
short int audio_format;
short int num_channels;
int sample_rate;
int byte_rate;
short int block_align;
short int bits_per_sample;
short int extra_param_size;
char subchunk2_id[4];
int subchunk2_size;
} header;
typedef struct header* header_p;
void scale_wav_file(char * input, float factor, int is_8bit)
{
FILE * infile = fopen(input, "rb");
FILE * outfile = fopen("outfile.wav", "wb");
int BUFSIZE = 4000, i, MAX_8BIT_AMP = 255, MAX_16BIT_AMP = 32678;
// used for processing 8-bit file
unsigned char inbuff8[BUFSIZE], outbuff8[BUFSIZE];
// used for processing 16-bit file
short int inbuff16[BUFSIZE], outbuff16[BUFSIZE];
// header_p points to a header struct that contains the file's metadata fields
header_p meta = (header_p)malloc(sizeof(header));
if (infile)
{
// read and write header data
fread(meta, 1, sizeof(header), infile);
fwrite(meta, 1, sizeof(meta), outfile);
while (!feof(infile))
{
if (is_8bit)
{
fread(inbuff8, 1, BUFSIZE, infile);
} else {
fread(inbuff16, 1, BUFSIZE, infile);
}
// scale amplitude for 8/16 bits
for (i=0; i < BUFSIZE; ++i)
{
if (is_8bit)
{
outbuff8[i] = factor * inbuff8[i];
if ((int)outbuff8[i] > MAX_8BIT_AMP)
{
outbuff8[i] = MAX_8BIT_AMP;
}
} else {
outbuff16[i] = factor * inbuff16[i];
if ((int)outbuff16[i] > MAX_16BIT_AMP)
{
outbuff16[i] = MAX_16BIT_AMP;
} else if ((int)outbuff16[i] < -MAX_16BIT_AMP) {
outbuff16[i] = -MAX_16BIT_AMP;
}
}
}
// write to output file for 8/16 bit
if (is_8bit)
{
fwrite(outbuff8, 1, BUFSIZE, outfile);
} else {
fwrite(outbuff16, 1, BUFSIZE, outfile);
}
}
}
// cleanup
if (infile) { fclose(infile); }
if (outfile) { fclose(outfile); }
if (meta) { free(meta); }
}
int main (int argc, char const *argv[])
{
char infile[] = "file.wav";
float factor = 0.5;
scale_wav_file(infile, factor, 0);
return 0;
}
我在最后得到不同的文件大小(以1K左右,对于一个40MB的文件),我怀疑这是由于这样的事实,我正在写一整个缓冲区输出,即使该文件在填充整个缓冲区大小之前可能已经终止。另外,输出文件会搞砸 - 不会播放或打开 - 所以我可能会做错整个事情。任何关于我搞砸的提示都会很棒。谢谢!
当你说的输入和输出文件具有不同的大小,是输出文件大于还是小于输入? – bta 2010-03-16 19:51:21
输出较大 – sa125 2010-03-16 20:07:35