Werner Flamme said the following on 07/23/2013 02:45 PM:
If you have to add 2 and 3 it will take the same time on both archs. 64 bit is not faster, it is larger. Compare a road truck of 6 wheels to another of 12 wheels and double length.
Huh? That doesn't make sense to me? Can you explain it it detail - and drop the analogy. Give op-code example. Doesn't have to be from a real machine, just hypothetical.
If you define 64bit integer variables and you store the values 2 and 3
... those are small integers ...
inside those, the 64bit math will be a faster, since it requires only one register operation (per variable). On 32bit architecture, it may take two - unless you have a compiler that is smart enough to check for zeroes before adding ;-)
Again, that doesn't make any sense. What zeros? Yes, 64-bit math will be faster, that is math involving integers greater than 2^32. How often do we do that? Apart, that is, from calculating addresses when we have an address space greater than can be addresses with 32-bits. But surely you're not doing that 'as arithmetic' but letting the compiler figure out the code for a pointer into an array. And I hope that you're not dealing with arrays of more than 2^32 elements ... What's that? Large text strings? You mean 'books'? -- The Soft overcomes the Hard The Slow overcomes the Fast Let your workings remain a mystery Just show people the Results -- To unsubscribe, e-mail: opensuse+unsubscribe@opensuse.org To contact the owner, e-mail: opensuse+owner@opensuse.org