function sha256.encrypt(s)
--initialize hash values
local h0 = 0x6a09e667
local h1 = 0xbb67ae85
local h2 = 0x3c6ef372
local h3 = 0xa54ff53a
local h4 = 0x510e527f
local h5 = 0x9b05688c
local h6 = 0x1f83d9ab
local h7 = 0x5be0cd19
--initialize round constants
local k = {0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5, 0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3, 0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc, 0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7, 0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13, 0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3, 0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5, 0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208, 0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2}
--preprocess string
local message = preprocessMessage(s)
--main compression
for i = 1, #message / 512 do
local w = createMessageSchedule(string.sub(message, 512 * (i - 1) + 1, 512 * (i - 1) + 512))
--initialize working values
local a = h0
local b = h1
local c = h2
local d = h3
local e = h4
local f = h5
local g = h6
local h = h7
for i = 1, 64 do
local S1 = bit.bxor(bit.bxor(xbit.brrotate(e, 6), xbit.brrotate(e, 11)), xbit.brrotate(e, 25))
local ch = bit.bxor(bit.band(e, f), bit.band(bit.bnot(e), g))
local temp1 = h + S1 + ch + k[i] + w[i]
local S0 = bit.bxor(bit.bxor(xbit.brrotate(a, 2), xbit.brrotate(a, 13)), xbit.brrotate(a, 22))
local maj = bit.bxor(bit.bxor(bit.band(a, b), bit.band(a, c)), bit.band(b, c))
local temp2 = S0 + maj
h = g
g = f
f = e
e = d + temp1
d = c
c = b
b = a
a = temp1 + temp2
h0 = (h0 + a) % 2^32
h1 = (h1 + b) % 2^32
h2 = (h2 + c) % 2^32
h3 = (h3 + d) % 2^32
h4 = (h4 + e) % 2^32
h5 = (h5 + f) % 2^32
h6 = (h6 + g) % 2^32
h7 = (h7 + h) % 2^32
end
end
return string.format('%08x%08x%08x%08x%08x%08x%08x%08x', h0, h1, h2, h3, h4, h5, h6, h7)
end
I am here trying to create my own implementation of SHA-256 for ComputerCraft. Yes, I know this is a pointless endeavor and I could just grab a prewritten library written by a more skilled programmer for this purpose, but there's no fun in doing that. I've gotten it to a state where it yields deterministic (same input = same output every time), fairly unpredictable (you can't easily tell what it will output based on the input), 64-byte-wide string outputs, but the output is still different from examples, telling me I did something wrong somewhere along the way.
tips: band = bitwise and, bor = bitwise or, bxor = bitwise xor (these are ccraft library functions)
I just need a fresh pair of eyes because I am sick of looking at this trying to figure out what went wrong. I've already verified that the function definitions I haven't included (preprocessMessage, createMessageSchedule) are working as intended.
Actual output for ''
d877d0451a38a0ef791326d2bfd61ab65b2348741f88c19b757670864c457e2f
Expected output for ''
e3b0c44298fc1c149afbf4c8996fb92427ae41e4649b934ca495991b7852b855
Actual output for 'hello world'
7c892f56e8ad9c894d37007889c7964d346f54875bfe1faf9c33e0f8870f9010
Expected output for 'hello world'
b94d27b9934d3e08a52e52d7da7dabfac484efe37a5380ee9088f7ace2efcde9
I am trying to use julai as main language for my work. But I find that this plot is different than python (Which outputs the right plot)
Here is the python code and output
import numpy as np
import math
import matplotlib.pyplot as plt
u = 9.27*10**(-21)
k = 1.38*10**(-16)
j2 = 7/2
nrr = 780
h = 1000
na = 6*10**(23)
rho = 7.842
mgd = 157.25
a = mgd
d = na*rho*u/a
m_f = []
igd = 7.0
for t in range(1,401):
while True:
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = abs(7*(q2*math.tanh(q2*x2)**-1 - (1/x4)*math.tanh(x2/x4)**-1))
if abs(m - igd) < 10**(-12):
break
else:
igd = m
m_f.append(abs(m))
plt.plot(range(1,401), m_f)
plt.savefig("Py_plot.pdf")
and it gives the following right plot
The right plot as expected
But when I do the same calculations in julia it gives different output than python, here is my julia code
using Plots
u = 9.27*10^(-21)
k = 1.38*10^(-16)
j2 = 7/2
nrr = 780
h = 1000
na = 6*10^(23)
rho = 7.842
mgd = 157.25
a = mgd
d = na*rho*u/a
igd = 7.0
m = 0.0
m_f = Float64[]
for t in 1:400
while true
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = 7*(q2*coth(rad2deg(q2*x2))-(1/x4)*coth(rad2deg(x2/x4)))
if abs(abs(m)-igd) < 10^(-10)
break
else
igd = m
end
end
push!(m_f, abs(m))
end
plot(1:400, m_f)
and this is the unexpected julia output
unexpected wrong output from julia
I wish for help....
Code:
using Plots
const u = 9.27e-21
const k = 1.38e-16
const j2 = 7/2
const nrr = 780
const h = 1000
const na = 6.0e23
const rho = 7.842
const mgd = 157.25
const a = mgd
const d = na*rho*u/a
function plot_graph()
igd = 7.0
m = 0.0
trange = 1:400
m_f = Vector{Float64}(undef, length(trange))
for t in trange
while true
h1 = h+d*nrr*igd
x2 = (7*u*h1)/(k*t)
x4 = 2*j2
q2 = (x4+1)/x4
m = abs(7*(q2*coth(q2*x2)-(1/x4)*coth(x2/x4)))
if isapprox(m, igd, atol = 10^(-10))
break
else
igd = m
end
end
m_f[t] = m
end
plot(trange, m_f)
end
Plot:
Changes for correctness:
Changed na = 6.0*10^(23) to na = 6.0e23.
Since ^ has a higher precedence than *, 10^23 is evaluated first, and since the operands are Int values, the result is also an Int. However, Int (i.e. Int64) can only hold numbers up to approximately 9 * 10^18, so 10^23 overflows and gives a wrong result.
julia> 10^18
1000000000000000000
julia> 10^19 #overflow starts here
-8446744073709551616
julia> 10^23 #and gives a wrong value here too
200376420520689664
6.0e23 avoids this problem by directly using the scientific e-notation to create a literal Float64 value (Float64 can hold this value without overflowing).
Removed the rad2deg calls when calling coth. Julia trigonometric functions by default take radians, so there's no need to make this conversion.
Other changes
Marked all the constants as const, and moved the rest of the code into a function. See Performance tip: Avoid non-constant global variables
Changed the abs(m - igd) < 10^-10 to isapprox(m, igd, atol = 10^-10) which performs basically the same check, but is clearer and more flexible (for eg. if you wanted to change to a relative tolerance rtol later).
Stored the 1:400 as a named variable trange. This is just because it's used multiple times, so it's easier to manage as a variable.
Changed m_f = Float64[] to m_f = Vector{Float64}(undef, length(trange)) (and the push! at the end to an assignment). If the size of the array is known beforehand (as it is in this case), it's better for performance to pre-allocate it with undef values and then assign to it.
Changed u and k also to use the scientific e-notation, for consistency and clarity (thanks to #DNF for suggesting the use of this notation in the comments).
Help me I need to create a Userform for my personal calculations. But I ran into a lot of problems. Because I've never written a program before.
When I enter a value 0 or enter a value other than 0 and delete it in the text field PriceCoinBuy, BuyCoin , PriceCoinSell , SellCoin , Vat one of the fields I will get Msg : Run-time error '6'; overflow.
But when I put a number other than 0 in the BuyCoin field, I get Msg : Run-time error '11'; Division by zero.
I've been searching for a solution for 2 days but can't find it in please help.
I have 5 textboxes for input.
Sub SumAll()
Dim A, B, C, D, E, F, G, H, I, J, K, V As Double
A = Val(Order.PriceCoinBuy.Value)
B = Val(Order.BuyCoin.Value)
C = Val(Order.PriceCoinSell.Value)
D = Val(Order.SellCoin.Value)
V = Val(Order.Vat.Value)
'-------------- Math --------------
E = CDbl(B) / A
F = CDbl(E) * (V / 100)
G = CDbl(E) - F
H = CDbl(G) * A
I = CDbl(D) * C
J = CDbl(I) * (V / 100)
K = CDbl(I) - J
'---------------- Show -------------
Order.GetCoin.Text = Format(E, "##,##0.000000")
Order.AfterVatBuy.Text = Format(F, "##,##0.0000")
Order.CoinBalance.Text = Format(G, "##,##0.000000")
Order.ToMoney.Text = Format(H, "##,##0.00")
Order.GetMoney.Text = Format(I, "##,##0.00")
Order.AfterVatSell.Text = Format(J, "##,##0.00")
Order.MoneyBalance.Text = Format(K, "##,##0.00")
End Sub
I'm a beginner in Fortran. I have to make a program with a function and an array but it doesn't work. I tried to use debugger :
-g -fcheck-bounds -fbacktrace -fimplicit-none etc
but nothing is wrong.
Here is my function :
FUNCTION tabspot (i, j, OFFSHORE, hauteurhoulemin, hauteurhoulemax)
IMPLICIT NONE
!---------------------------------------
! Déclaration des arguments
!---------------------------------------
INTEGER, INTENT(INOUT) :: i, j
CHARACTER(LEN=5), INTENT(OUT) :: OFFSHORE
INTEGER, INTENT(OUT):: hauteurhoulemin
INTEGER, INTENT(OUT):: hauteurhoulemax
!---------------------------------------
! Déclaration du type de la fonction
!---------------------------------------
CHARACTER(LEN=14), DIMENSION(4) :: tabspot
!---------------------------------------
! Déclaration des varibles locales
!---------------------------------------
CHARACTER(LEN=14) :: box
tabspot(1) = 'Cotedesbasques'
tabspot(2) = 'Mundaka'
tabspot(3) = 'Parlementia'
tabspot(4) = 'Hendaye'
DO i=1,4
SELECT CASE (i)
CASE(1)
OFFSHORE = 'EST'
hauteurhoulemin = 30
hauteurhoulemax = 150
CASE(2)
OFFSHORE = 'SUD'
hauteurhoulemin = 90
hauteurhoulemax = 400
CASE(3)
OFFSHORE = 'EST'
hauteurhoulemin = 120
hauteurhoulemax = 500
CASE(4)
OFFSHORE = 'SUD'
hauteurhoulemin = 60
hauteurhoulemax = 200
END SELECT
DO j=i+1,5
IF (((vent1%direction == OFFSHORE) .AND. (hauteurhoule>=hauteurhoulemin)&
.AND. (hauteurhoule<=hauteurhoulemax)) .OR. &
((vent1%force == 0) .AND. (hauteurhoule>=hauteurhoulemin)&
.AND. (hauteurhoule<=hauteurhoulemax)))THEN
box=tabspot(i)
tabspot(i)=tabspot(j)
tabspot(j)=box
END IF
END DO
END DO
END FUNCTION
Thank you in advance for your answer.
Martin
I have a string of alphabets e.g.
S = ['a';'b';'c';'d';'e'];
and I want to put it in column 3 in the table:
table(:,1) = M1; table(:,2) = d1;
disp(table)
M1 and d1 are 5 X 1 matrices of numbers each.
You could think to do as follows:
r = {M1, d1, S};
or
r = {M1; d1; S};
EDIT
you may also do this:
M1 = rand(5,1);
d1 = rand(5,1);
S = ['a';'b';'c';'d';'e'];
y = arrayfun(#(i) {M1(i), d1(i), S(i)'},1:5,'UniformOutput',false);
res = cat(1,y{:});
I'd suggest this approach to solve your question.