C++ 如何简化DirectX11上的矩阵乘法？

我是否可以简化以下语句，使其更具可读性？关于对齐要求，我可以一步一步地执行以下乘法吗

DirectX::XMStoreFloat4x4(&this->worldTransform, DirectX::XMMatrixMultiply(DirectX::XMMatrixMultiply(DirectX::XMMatrixScalingFromVector(DirectX::XMLoadFloat4(&this->scaling)), DirectX::XMMatrixRotationQuaternion(DirectX::XMLoadFloat4(&this->rotation))), DirectX::XMMatrixTranslationFromVector(DirectX::XMLoadFloat4(&this->translation))));

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使用

XMVECTOR

和

XMMATRIX

局部变量。在优化的构建中，生成的代码应该是相同的

XMVECTOR vTrans = DirectX::XMLoadFloat4(&this->translation);
XMMATRIX mTrans = DirectX::XMMatrixTranslationFromVector(vTrans);
XMVECTOR vScale = DirectX::XMLoadFloat4(&this->scaling);
XMMATRIX mScale = DirectX::XMMatrixScalingFromVector(vScale);
XMVECTOR vRot = DirectX::XMLoadFloat4(&this->rotation);
XMMATRIX mRot = DirectX::XMMatrixRotationQuaternion(rot);
XMMATRIX mTemp = DirectX::XMMatrixMultiply(mScale, mRot);
mTemp = DirectX::XMMatrixMultiply(mTemp, mTrans);
DirectX::XMStoreFloat4x4(&this->worldTransform, mTemp);