9 ms·

I Updated the run bench function https://gist.github.com/mzaks/e3a2dc7ccdfc2397bc26c55eb6dc8ac3 https://gist.github.com/mzaks/e3a2dc7ccdfc2397bc26c55eb6dc8a...

by mzaks 10y ago

I Updated the run bench function https://gist.github.com/mzaks/e3a2dc7ccdfc2397bc26c55eb6dc8ac3 https://gist.github.com/mzaks/e3a2dc7ccdfc2397bc26c55eb6dc8a... the output is now: Eager run ================================= 1557 ms encode 264 ms decode 34 ms use 206 ms dealloc 504 ms decode+use+dealloc 0,38 ms direct 0,32 ms using struct ================================= Total counter1 is 8644311667000000 Total counter2 is 8644311667000000 Total counter3 is 8644311667000000 Encoded size is 315 bytes, should be 344 if not using unique strings ================================= As you can see all three counters are equal.

deleted 10y ago
[deleted]
toth 10y ago
The function call is not being optimized out, it's being hoisted outside the loop. I.e., it is as if the code was written as: let result = flatuseStruct(outputData) for _ in 0..<iterations { assert(result == 8644311667) total = total + UInt64(result) } The counter will still be correct, but you are not measuring what you think you are measuring.
- mzaks 10y ago
  This makes sense! Changed the iteration to: for i in 0..<iterations { let result = flatuseStruct(outputData, start:i) assert(result == 8644311666 + Int(i)) total2 = total2 + UInt64(result) } now result is around 43ms Thanks for pointing it out. Have to check if there are so other things involved, but that might be it.
to3m 10y ago
You can see the problem here: First it cals CFAbsoluteTimeGetCurrent and saves the result. 0x100272dab <+10443>: callq 0x1002b7b38 ; CFAbsoluteTimeGetCurrent 0x100272db0 <+10448>: movapd %xmm0, -0xa0(%rbp) Here is the call to flatDecodeDirect. I guess RDI is the input. That's usual for the x64 ABI. 0x100272db8 <+10456>: movq -0x100(%rbp), %rdi 0x100272dbf <+10463>: callq 0x10026fb10 ; flatDecodeDirect I don't know what this next bit is for. 0x100272dc4 <+10468>: testq %rax, %rax 0x100272dc7 <+10471>: js 0x10027444c ; <+16236> [inlined] generic specialization <FlatBuffersPerformanceTestDesktop.FlatBufferReader> of Swift._ContiguousArrayBuffer._checkValidSubscript (Swift.Int) -> () 0x3e8=1000. The loop counter is in ECX. 0x100272dcd <+10477>: movl $0x3e8, %ecx I can't figure out what's at these two addresses; lldb didn't seem to accept any reasonable syntax. lldb is terrible. But I'll bet that RBX is holding the value of `total'. I don't know what r14 is, and it doesn't seem to matter since nothing here uses it. 0x100272dd2 <+10482>: movq -0xd8(%rbp), %rbx 0x100272dd9 <+10489>: movq -0x198(%rbp), %r14 Here's the loop. The loop is unrolled 5 times. total+=result. 0x100274454 produces some kind of exception on integer overflow. 0x100272de0 <+10496>: addq %rax, %rbx 0x100272de3 <+10499>: jb 0x100274454 ; at flatbench.swift:284 0x100272de9 <+10505>: addq %rax, %rbx 0x100272dec <+10508>: jb 0x100274454 0x100272df2 <+10514>: addq %rax, %rbx 0x100272df5 <+10517>: jb 0x100274454 0x100272dfb <+10523>: addq %rax, %rbx 0x100272dfe <+10526>: jb 0x100274454 0x100272e04 <+10532>: addq %rax, %rbx 0x100272e07 <+10535>: jb 0x100274454 The loop was unrolled 5 times, so drop 5 from the loop counter and repeat. 0x100272e0d <+10541>: addq $-0x5, %rcx 0x100272e11 <+10545>: jne 0x100272de0 ; at flatbench.swift:276 Get current time. 0x100272e13 <+10547>: callq 0x1002b7b38 ; CFAbsoluteTimeGetCurrent So this code actually times one call to flatDecodeDirect, then 200 iterations of an unrolled do-nothing loop. The compiler has figured out somehow that flatDecodeDirect is going to do exactly the same thing each time, and taken advantage of that by calling it only once. I'm guessing this means that flatDecodeDirect is only called 1,000 times in total. As a sanity check for this kind of thing - try making a little loop that just increments an integer the appropriate number of times, and see how long that takes. (Check the assembly language output to ensure the generated code is doing what you think - it should be a 2-instruction loop.) On my laptop that takes 1.8ms. This isn't the absolute limit of how long it takes to do 1,000,000 of anything, but it'll do as a rough estimate. So you should be suspicious if a program suggests it's taking much less time than that to do 1,000,000 of something that's a lot more complicated, as the test did. (It reported 1,000,000 iterations in 0.53ms on my PC.) (Of course, as with any rough estimate, this only gives you a suspicion, and isn't proof without further investigation.)
- mzaks 10y ago
  I was suspicious, I just could not put my finger on it :) for i in 0..<iterations { let result = flatuseStruct(outputData, start:i) assert(result == 8644311666 + Int(i)) total2 = total2 + UInt64(result) } this results in around 42ms compared to C 25ms. I guess I should update my blog post :) Thanks for your help.
easytiger 10y ago
That's 0.35 nanoseconds to do something. If it is a nop on a modern cpu it can run x4 nops at the same time. Either you aren't doing useful work or your measurements are wrong. Unfortunately I can't view gists in work.
deleted 10y ago
[deleted]

by mzaks 10y ago

deleted 10y ago

[deleted]

toth 10y ago

The function call is not being optimized out, it's being hoisted outside the loop. I.e., it is as if the code was written as: let result = flatuseStruct(outputData) for _ in 0..<iterations { assert(result == 8644311667) total = total + UInt64(result) } The counter will still be correct, but you are not measuring what you think you are measuring.

mzaks 10y ago
This makes sense! Changed the iteration to: for i in 0..<iterations { let result = flatuseStruct(outputData, start:i) assert(result == 8644311666 + Int(i)) total2 = total2 + UInt64(result) } now result is around 43ms Thanks for pointing it out. Have to check if there are so other things involved, but that might be it.

to3m 10y ago

You can see the problem here: First it cals CFAbsoluteTimeGetCurrent and saves the result. 0x100272dab <+10443>: callq 0x1002b7b38 ; CFAbsoluteTimeGetCurrent 0x100272db0 <+10448>: movapd %xmm0, -0xa0(%rbp) Here is the call to flatDecodeDirect. I guess RDI is the input. That's usual for the x64 ABI. 0x100272db8 <+10456>: movq -0x100(%rbp), %rdi 0x100272dbf <+10463>: callq 0x10026fb10 ; flatDecodeDirect I don't know what this next bit is for. 0x100272dc4 <+10468>: testq %rax, %rax 0x100272dc7 <+10471>: js 0x10027444c ; <+16236> [inlined] generic specialization <FlatBuffersPerformanceTestDesktop.FlatBufferReader> of Swift._ContiguousArrayBuffer._checkValidSubscript (Swift.Int) -> () 0x3e8=1000. The loop counter is in ECX. 0x100272dcd <+10477>: movl $0x3e8, %ecx I can't figure out what's at these two addresses; lldb didn't seem to accept any reasonable syntax. lldb is terrible. But I'll bet that RBX is holding the value of `total'. I don't know what r14 is, and it doesn't seem to matter since nothing here uses it. 0x100272dd2 <+10482>: movq -0xd8(%rbp), %rbx 0x100272dd9 <+10489>: movq -0x198(%rbp), %r14 Here's the loop. The loop is unrolled 5 times. total+=result. 0x100274454 produces some kind of exception on integer overflow. 0x100272de0 <+10496>: addq %rax, %rbx 0x100272de3 <+10499>: jb 0x100274454 ; at flatbench.swift:284 0x100272de9 <+10505>: addq %rax, %rbx 0x100272dec <+10508>: jb 0x100274454 0x100272df2 <+10514>: addq %rax, %rbx 0x100272df5 <+10517>: jb 0x100274454 0x100272dfb <+10523>: addq %rax, %rbx 0x100272dfe <+10526>: jb 0x100274454 0x100272e04 <+10532>: addq %rax, %rbx 0x100272e07 <+10535>: jb 0x100274454 The loop was unrolled 5 times, so drop 5 from the loop counter and repeat. 0x100272e0d <+10541>: addq $-0x5, %rcx 0x100272e11 <+10545>: jne 0x100272de0 ; at flatbench.swift:276 Get current time. 0x100272e13 <+10547>: callq 0x1002b7b38 ; CFAbsoluteTimeGetCurrent So this code actually times one call to flatDecodeDirect, then 200 iterations of an unrolled do-nothing loop. The compiler has figured out somehow that flatDecodeDirect is going to do exactly the same thing each time, and taken advantage of that by calling it only once. I'm guessing this means that flatDecodeDirect is only called 1,000 times in total. As a sanity check for this kind of thing - try making a little loop that just increments an integer the appropriate number of times, and see how long that takes. (Check the assembly language output to ensure the generated code is doing what you think - it should be a 2-instruction loop.) On my laptop that takes 1.8ms. This isn't the absolute limit of how long it takes to do 1,000,000 of anything, but it'll do as a rough estimate. So you should be suspicious if a program suggests it's taking much less time than that to do 1,000,000 of something that's a lot more complicated, as the test did. (It reported 1,000,000 iterations in 0.53ms on my PC.) (Of course, as with any rough estimate, this only gives you a suspicion, and isn't proof without further investigation.)

mzaks 10y ago
I was suspicious, I just could not put my finger on it :) for i in 0..<iterations { let result = flatuseStruct(outputData, start:i) assert(result == 8644311666 + Int(i)) total2 = total2 + UInt64(result) } this results in around 42ms compared to C 25ms. I guess I should update my blog post :) Thanks for your help.

easytiger 10y ago

That's 0.35 nanoseconds to do something. If it is a nop on a modern cpu it can run x4 nops at the same time. Either you aren't doing useful work or your measurements are wrong. Unfortunately I can't view gists in work.

deleted 10y ago

[deleted]