CODEWITHSUNDEEP
graphicsdirectxdirect3d
cecil
cecil

Reputation: 453

Allocating memory for a texture

I've been looking at the WICTextureLoader.cpp file that is part of the DirectX Toolkit. And I'm trying to understand this bit of code:

https://github.com/microsoft/DirectXTK/blob/master/Src/WICTextureLoader.cpp#L530

   // Allocate temporary memory for image
   uint64_t rowBytes = (uint64_t(twidth) * uint64_t(bpp) + 7u) / 8u;
   uint64_t numBytes = rowBytes * uint64_t(theight);

I understand that you'd need to allocate the texture width * bits per pixel for a single row of the texture. But I do not understand the addition of 7 per pixel and then the division of the whole thing by 8. What is that about?

Upvotes: 0

Views: 182

Answers (1)

Chuck Walbourn
Chuck Walbourn

Reputation: 41047

bpp here is "bits per pixel". The expression rounds up to the next whole byte (8 bits).

This is a classic C pattern for "align up" for power-of-2 alignments, here expressed as a C++ template.

    template<typename T>
    inline T AlignUp(T size, size_t alignment) noexcept
    {
        if (alignment > 0)
        {
            assert(((alignment - 1) & alignment) == 0);
            auto mask = static_cast<T>(alignment - 1);
            return (size + mask) & ~mask;
        }
        return size;
    }

Instead of / 8, I could have used bit operators (again, since it's a power of two) and done & ~8 but the / 8 seems clearer.

Upvotes: 1

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