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Transistors, I guess; HT adds extra registers for each core. However, I have some doubts that removing them would make a tangible impact on the final price.
by pizza234 8y ago
Transistors, I guess; HT adds extra registers for each core. However, I have some doubts that removing them would make a tangible impact on the final price.
- kingosticks 8y agoOr you could leave it all in there (cheaper and easier) just disabled and instead enjoy better yields: * Parts with broken HT can be shipped as fully working non-HT parts. * High-leakage parts can be shipped as they now pass the power screening.
- gsnedders 8y agoThis is the status quo with i5 v. i7 binning, AFAIK.
- wtallis 8y ago> Parts with broken HT can be shipped as fully working non-HT parts. It's unlikely that they ever get parts with broken HT on an otherwise salvageable core. Too much of that functionality is simply partitioning existing resources in half. There aren't that many transistors that are simply HT overhead that the core can do without when HT is not in use.
- theandrewbailey 8y agoIntel claimed that HT used about 5% of the die area, but that was with HT Pentium 4. The numbers probably changed in 15 years.
- monocasa 8y agoIt's probably less as a percentage now.
- Symmetry 8y agoNot registers per se. You want a large number of physical registers for renaming and speculative execution in any event. What you need with HT is another sets of mappings form architectural to physical registers for the committed state.
- dragontamer 8y agoNot actually. Remember that architectural registers (such as RAX or EAX) are "fake" and remapped to the real, physical, microcode registers. Code like "xor eax, eax" is translated into "physical-register #51 = 0" with regards to uops, and doesn't even use any execution units on modern Skylake or Zen processors! Hyperthreads only need to contain one more remapping of physical registers to architectural registers.