3 ms·

What about type conversion? The values were `uint32_t`, but converted to `uint64_t`, doesn't that consume cycles as well, or for one operation, optimizer compil

• wizzwizz4 5y ago
  Some CPUs have a single instruction for “multiply but get the high word”.
• flohofwoe 5y ago
  Multiplying two 32-bit numbers yields a 64-bit result down on the CPU instruction level, so the type conversion is basically built in, details may differ by CPU, but here's the vanilla x86 multiply which spreads the results over two registers: https://c9x.me/x86/html/file_module_x86_id_210.html https://c9x.me/x86/html/file_module_x86_id_210.html ...so the compiler could even drop the right-shift and instead just use the register where the upper half of the result is stored.
  • deepsun 5y ago
    Interesting, thank you. But the type conversion is on the results, not input. Is it guaranteed that we're always multiplying two 32-bit numbers and not two 64-bit numbers in "(uint64_t) x * (uint64_t) N"? Because looking at the code, it's more natural to think we're multiplying 64-bit integers, so we may need some CPU cycles to make "x" occupy 64 bits first, same for "N", and then do multiplication on them.
    • pwuille 5y ago
      The conversion is free, at least on x86-like platforms, because there are no separate 32-bit and 64-bit registers. Instead, there is a fixed shared set of registers, and the instructions signal whether they operate on the 32-bit or 64-bit value on it. When assigning to a register in 32-bit mode the top 32 bits are cleared, so if x and n were loaded into 32-bit registers, a 64-bit multiplication can be applied to it directly, without any conversion instructions.