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by sifar 1136 days ago
Generally, since the memory is byte addressable, you load data which is packed into bytes. It is the compute instructions that use the specified bits needed.

So in this case one would load a byte which would have 2 4b data, and then you would have a 4b ADD or MAC which would operate on them.

If you don't have them then you need to sign/zero extend or convert the smaller bit-widths to 8/16/32b whichever is available.