Trying to read binary file into struct with julia

When working with Julia, you may come across the need to read binary files into a struct. This can be a bit tricky, but there are several ways to accomplish it. In this article, we will explore three different approaches to solve this problem.

Approach 1: Using the `reinterpret` function

One way to read a binary file into a struct is by using the `reinterpret` function. This function allows you to reinterpret the binary data as a different type. Here’s an example:

struct MyStruct
    field1::Int32
    field2::Float64
end

function read_binary_file(filename::String)
    data = read(filename, UInt8)
    num_bytes = sizeof(MyStruct)
    num_elements = length(data) ÷ num_bytes
    result = Vector{MyStruct}(undef, num_elements)
    
    for i in 1:num_elements
        offset = (i-1) * num_bytes + 1
        result[i] = reinterpret(MyStruct, data[offset:offset+num_bytes-1])
    end
    
    return result
end

In this approach, we first read the binary file as an array of `UInt8` using the `read` function. Then, we calculate the number of elements in the file by dividing the length of the data by the size of the struct. We create a vector of `MyStruct` objects with the calculated number of elements. Finally, we use the `reinterpret` function to reinterpret the binary data as `MyStruct` and assign it to the corresponding element in the result vector.

Approach 2: Using the `read!` function

Another approach is to use the `read!` function, which allows you to read binary data directly into a preallocated buffer. Here’s an example:

struct MyStruct
    field1::Int32
    field2::Float64
end

function read_binary_file(filename::String)
    num_bytes = sizeof(MyStruct)
    num_elements = filesize(filename) ÷ num_bytes
    result = Vector{MyStruct}(undef, num_elements)
    
    open(filename, "r") do file
        buffer = Vector{UInt8}(undef, num_bytes)
        
        for i in 1:num_elements
            read!(file, buffer)
            result[i] = reinterpret(MyStruct, buffer)
        end
    end
    
    return result
end

In this approach, we first calculate the number of elements in the file by dividing the file size by the size of the struct. We create a vector of `MyStruct` objects with the calculated number of elements. Then, we open the file and create a buffer of `UInt8` with the size of the struct. We use the `read!` function to read binary data from the file into the buffer, and then reinterpret it as `MyStruct` and assign it to the corresponding element in the result vector.

Approach 3: Using the `Mmap.mmap` function

A third approach is to use the `Mmap.mmap` function from the `Mmap` package. This function allows you to memory-map a file, which means that you can access its contents as if they were in memory. Here’s an example:

using Mmap

struct MyStruct
    field1::Int32
    field2::Float64
end

function read_binary_file(filename::String)
    num_bytes = sizeof(MyStruct)
    num_elements = filesize(filename) ÷ num_bytes
    result = Vector{MyStruct}(undef, num_elements)
    
    open(filename, "r") do file
        mmap_data = Mmap.mmap(file)
        
        for i in 1:num_elements
            offset = (i-1) * num_bytes + 1
            result[i] = reinterpret(MyStruct, mmap_data[offset:offset+num_bytes-1])
        end
    end
    
    return result
end

In this approach, we first calculate the number of elements in the file by dividing the file size by the size of the struct. We create a vector of `MyStruct` objects with the calculated number of elements. Then, we open the file and use the `Mmap.mmap` function to memory-map it. This allows us to access the file’s contents as if they were in memory. We then use the same logic as in the previous approaches to reinterpret the binary data as `MyStruct` and assign it to the corresponding element in the result vector.

After exploring these three approaches, it is clear that the best option depends on the specific requirements of your project. If memory efficiency is a concern, Approach 1 using `reinterpret` may be the most suitable. If you prefer a more straightforward and readable code, Approach 2 using `read!` is a good choice. Finally, if you need to work with very large files and want to minimize disk I/O, Approach 3 using `Mmap.mmap` can be the most efficient.

Ultimately, the best option is the one that meets your project’s requirements in terms of performance, memory usage, and code readability.