如何以“ 4字节单个” /浮点数/ IEEE 754编码数据的二进制数据读取ArrayBuffer？

Question

I need to loop over a binary file via an arrayBuffer and retrieve sets of 1024 floating points. I'm doing this:

// chunk_size = 1024
// chunk_len = 48
// response_buffer = ArrayBuffer
// response_buffer.byteLength = 49152
for (i = chunk_len; i > 0; i -= 1) {
    switch (i) {
        case (chunk_len):

            // create view from last chunk of 1024 out of 49152 total
            float_buffer = new Float32Array(response_buffer, ((i - 1) * chunk_size));

            // add_data(net_len, float_buffer);
            break;

        case 0:
            break;

        default:
            float_buffer = new Float32Array(response_buffer, ((i - 1) * chunk_size)), chunk_size);
            //add_data(net_len, float_buffer);
            break;
    }
}

My problem is if I call this on the first run for the end of my buffer:

// i - 1 = 47 * chunk_size
new Float32Array(response_buffer, ((i - 1) * chunk_size));

the same statement fails on the next run where I call:

new Float32Array(response_buffer, ((i - 1) * chunk_size), 1024);

Although I can read here , that I can do this:

Float32Array Float32Array(
    ArrayBuffer buffer,
    optional unsigned long byteOffset,
    optional unsigned long length
);

Question :
Why is my loop failing after declaring the first Float32Array view on my response_offer ArrayBuffer?

Answer 1

I think you have an extra ")" in the first line of your "default" case.

float_buffer = new Float32Array(response_buffer, ((i - 1) * chunk_size)), chunk_size);

Should be:

float_buffer = new Float32Array(response_buffer, ((i - 1) * chunk_size), chunk_size);

Answer 2

So. Finally understand... maybe this helps the next person wondering:

First off - I was all wrong in trying to read my data, which is 4 byte single format .

• If I have an arrayBuffer with byteLength = 49182 this means there are that many entries in my array.
• Since my array is 4 byte single format , I found out with some SO-help and searching that this is readable with getFloat32 AND that 4 entries comprise one "real" value
• My data contains 3 measurements a 4000 data points stored in units of 1024 column by column.

• So if I have 12000 data-points and 49182/4 = 12288 datapoints, I will have 288 empty data points at the end of my data structure.

• So my binary data should be stored like this:

    0 - 1024 a 1025 - 2048 a 2049 - 3072 a 3072 - 4000 [a 4001 - 4096 b] 4097 - 5120 b 5121 - 6144 b 6145 - 7168 b 7169 - 8000 [b 8001 - 8192 c] 8193 - 9216 c 9217 - 10240 c 10240 - 11264 c 11264 - 12000 [c 12000 - 12288 0] 

• My final snippet will contain 288 empty results, because 3x4000 datapoints in 1024 chunks will return some empty results

• To read, I found a nice snippet here (dynamic high range rendering) , which helped me to this:

    // ... raw_data = ... data = new DataView(raw_data); ... tmp_data = new Float32Array(byte_len / Float32Array.BYTES_PER_ELEMENT); len = tmp_data.length; // Incoming data is raw floating point values with little-endian byte ordering. for (i = 0; i < len; i += 1) { tmp_data[i] = data.getFloat32(i * Float32Array.BYTES_PER_ELEMENT, true); } 

• Now I have a single array with which I can work and build my processing structure.