4 ms·

I just tried this on Python3 on a 64-bit x86 system: import math x = 0.49999999999999994 print(x-0.5) print(math.floor(x+0.5)) I got these pri

I just tried this on Python3 on a 64-bit x86 system: import math x = 0.49999999999999994 print(x-0.5) print(math.floor(x+0.5)) I got these printouts: -5.551115123125783e-17 1 So yes, something less than 1/2, with 1/2 added to it, has a floor of 1 in floating point math. Yet another reminder that floating point calculations are approximations, and not exact.

deleted 7y ago
[deleted]
e12e 7y ago
In Julia it's easy to see the difference between promoting to full number tower and sticking to 64 floats (ed: note difference in result 0 vs 1): julia> using Benchmarktools julia> @btime floor(0.49999999999999997+0.5) 0.027 ns (0 allocations: 0 bytes) 1.0 julia> @btime floor(0.49999999999999997+BigFloat(0.5)) 280.594 ns (6 allocations: 336 bytes) 0.0 julia> @code_native floor(0.49999999999999997+BigFloat(0.5)) .text ; ┌ @ floatfuncs.jl:152 within `floor' subq $24, %rsp movq %rsi, 16(%rsp) movq (%rsi), %rax ; │┌ @ floatfuncs.jl:152 within `#floor#543' movq %rax, (%rsp) movabsq $jl_system_image_data, %rax movq %rax, 8(%rsp) movabsq $japi1_round_16005, %rax movabsq $jl_system_image_data, %rdi movq %rsp, %rsi movl $2, %edx callq *%rax ; │└ addq $24, %rsp retq nopw %cs:(%rax,%rax) ; └ julia> @code_native floor(0.49999999999999997+0.5) .text ; ┌ @ floatfuncs.jl:152 within `floor' ; │┌ @ floatfuncs.jl:152 within `#floor#543' ; ││┌ @ floatfuncs.jl:152 within `round' vroundsd $9, %xmm0, %xmm0, %xmm0 ; │└└ retq nopw (%rax,%rax) ; └
masklinn 7y ago
> Yet another reminder that floating point calculations are approximations, and not exact. floating-point calculations are absolutely exact. What's not is conversions between decimals and floating-point.
- murkle 7y ago
  > floating-point calculations are absolutely exact Are you sure about that? What about 1.00000000000001^2 (using eg 64 bit double)?
  - lifthrasiir 7y ago
    Maybe the parent wanted to say that the result is detetministic and not far too off from the true value (in fact, as accurate as possible given the format). In the other words FP is exact but abides by a slight different calculation rule involving rounding.
    - dwheeler 7y ago
      Floating point is usually deterministic, but from a math point of view it is inexact (in most cases it only approximates the exact answer). In many cases that is fine, which is why it is used, but it is important to remember that.
- edflsafoiewq 7y ago
  Their not being exact is why 0.49999997 + 0.5 = 1. There isn't enough precision in the range just below 1.0 to represent the number 0.99999997.

deleted 7y ago

[deleted]

e12e 7y ago

In Julia it's easy to see the difference between promoting to full number tower and sticking to 64 floats (ed: note difference in result 0 vs 1): julia> using Benchmarktools julia> @btime floor(0.49999999999999997+0.5) 0.027 ns (0 allocations: 0 bytes) 1.0 julia> @btime floor(0.49999999999999997+BigFloat(0.5)) 280.594 ns (6 allocations: 336 bytes) 0.0 julia> @code_native floor(0.49999999999999997+BigFloat(0.5)) .text ; ┌ @ floatfuncs.jl:152 within `floor' subq $24, %rsp movq %rsi, 16(%rsp) movq (%rsi), %rax ; │┌ @ floatfuncs.jl:152 within `#floor#543' movq %rax, (%rsp) movabsq $jl_system_image_data, %rax movq %rax, 8(%rsp) movabsq $japi1_round_16005, %rax movabsq $jl_system_image_data, %rdi movq %rsp, %rsi movl $2, %edx callq *%rax ; │└ addq $24, %rsp retq nopw %cs:(%rax,%rax) ; └ julia> @code_native floor(0.49999999999999997+0.5) .text ; ┌ @ floatfuncs.jl:152 within `floor' ; │┌ @ floatfuncs.jl:152 within `#floor#543' ; ││┌ @ floatfuncs.jl:152 within `round' vroundsd $9, %xmm0, %xmm0, %xmm0 ; │└└ retq nopw (%rax,%rax) ; └

masklinn 7y ago

> Yet another reminder that floating point calculations are approximations, and not exact. floating-point calculations are absolutely exact. What's not is conversions between decimals and floating-point.

murkle 7y ago
> floating-point calculations are absolutely exact Are you sure about that? What about 1.00000000000001^2 (using eg 64 bit double)?
- lifthrasiir 7y ago
  Maybe the parent wanted to say that the result is detetministic and not far too off from the true value (in fact, as accurate as possible given the format). In the other words FP is exact but abides by a slight different calculation rule involving rounding.
  - dwheeler 7y ago
    Floating point is usually deterministic, but from a math point of view it is inexact (in most cases it only approximates the exact answer). In many cases that is fine, which is why it is used, but it is important to remember that.
edflsafoiewq 7y ago
Their not being exact is why 0.49999997 + 0.5 = 1. There isn't enough precision in the range just below 1.0 to represent the number 0.99999997.