C在运行时选择变量类型

Question

我在下面的这段代码看起来非常明确和冗余,有没有办法在运行时选择变量类型？

if(header->bitsPerSample == 16) {
    int sample;
    for (int i = 0; i < header->chunkSize - HEADER_SIZE + 8; i += sizeof(sample) * 2) {
        fread(&sample, sizeof(sample), 1, fp);
        fwrite(&sample, sizeof(sample), 1, new);
        fread(&sample, sizeof(sample), 1, fp);
    }
} else if(header->bitsPerSample == 8) {
    char sample;
    for (int i = 0; i < header->chunkSize - HEADER_SIZE + 8; i += sizeof(sample) * 2) {
        fread(&sample, sizeof(sample), 1, fp);
        fwrite(&sample, sizeof(sample), 1, new);
        fread(&sample, sizeof(sample), 1, fp);
    }
}

我正在寻找这样的东西:

if(header->bitsPerSample == 16) 
    sample is of type int
else if(header->bitsPerSample == 8)
    sample is of type char

for (int i = 0; i < header->chunkSize - HEADER_SIZE + 8; i += sizeof(sample) * 2) {
        fread(&sample, sizeof(sample), 1, fp);
        fwrite(&sample, sizeof(sample), 1, new);
        fread(&sample, sizeof(sample), 1, fp);
}

Answer 1

C编写通用代码的唯一工具是使用宏,但这不是必需的.

略过声明一个正确类型的变量的形式,而只是读取或写入适当的字节数:

size_t raw_size = header->bitsPerSample / 8;
unsigned char buffer[raw_size];
void* raw = &buffer;

for (int i = 0; i < header->chunkSize - HEADER_SIZE + 8; i += raw_size * 2) {
  fread(raw, raw_size, 1, fp);
  fwrite(raw, raw_size, 1, new);
  fread(raw, raw_size, 1, fp);
}

由于您的代码并不真正关心值是什么,只需要它们需要多少字节,您可以只使用字符缓冲区.