recovering double value from array of characters with iEEE 754

Question

So I am trying to read data from an STL file in 4 byte increments. Each 4 byte number corresponds to a value for an x,y, or z coordinate. For my purposes, the number doesn't need to be really precise. Storing the value as a double will do. However, to read the 4 byte number, I must store it initially in a char[4].

For example if I try to read in the number "2" i get my char[4] to look like 0 0 0 64

similarly, the number 0.314286 looks like 15 -22 -96 62

After looking at this link http://www.h-schmidt.net/FloatConverter/IEEE754.html I determined that these numbers correspond to a binary representation for IEEE754. In this format, the magnitude is still the number I want (the number I want being 2 or 0.314286 for example) but I can't get my program to correctly display those values.

for example: 0 0 0 64 = 00000000 00000000 00000000 01000000 = 2 in IEEE754 NOTE: this is slightly different than the website converter in the above link because I figured the endianness may be different, however the values are still correct.

How can I convert the char[4] of 4 bytes into the actual double value that it holds? Maybe this problem is a lot easier than I think it is but I seem to be having trouble casting the value.

I read into IEEE 754 on this site http://www.cprogramming.com/tutorial/floating_point/understanding_floating_point_representation.html but I'm not sure on what way I should approach this problem.

Thank you for reading this question, I am new to this community but I appreciate any direction I can get!

Here is a code sample of how I am reading the data from a file with name stored in fname.

ifstream myFile(fname.c_str(), ios::in | ios::binary);
char triData[4] = "";
myFile.read(triData, 4);

as for the input file, its a binary STL file. Therefore viewing the file doesn't make a ton of sense to us. Here is a snippet from a file for a sphere shape.

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Evan

Answer 1

Problem 1: The question reads as though OP is trying to read 4 bytes of data into a an 8 byte double. Solution to this is to use a 4 byte float, then assign to double if needed.

With data sizing out of the way, the quick solution is to read the data from the file directly into the float

float tridata;
if (inFile.read((char*)&tridata, sizeof(tridata))
{ 
    //TODO: consume tridata. May want to bounds-test tridata first.
}
else
{ 
    // TODO: Handle file IO error
}

This might not work depending on how the files were written and won't be portable across all platforms. It's most likely all that you need to do assuming the output files were generated on the same platform as the reader. It's worth trying this first to get things going.

Otherwise, read the data from the file into a generic byte buffer and convert to the local float byte ordering by swapping bytes.

Once the buffer has been ordered correctly, there are many ways to force the buffer to look like a float. Most of them will result undefined behaviour--they may work on processor and compiler X, but not Y and maybe not even reliably on X.

One thing that always works is the slow way: memcpy. With a modern compiler set to a high level of optimization it's typically not that slow.

//Open file
std::ifstream inFile(fname, std::ios::binary); // std::ios::in is implied by ifstream
// TODO: read to or seek to correct location in file
char buffer[sizeof(float)];
// read from file
if (inFile.read(buffer, sizeof(float))
{ // read 4 bytes of data from file
    float tridata;
    //TODO: flip endian as required
    std::memcpy(&tridata, buffer, sizeof(float));
    //TODO: consume tridata. May want to bounds-test tridata first.
}
else
{ // failed to read from file.
    // TODO: Handle file IO error
}