What reason is different result, when I change R code to Julia code? Thank you!

[meib...@163.com]

#####R code

f=c(0.5,0.5)

n.al <- length(f)

P.1 <- P.2 <- array(0,dim=rep(n.al,4))

    for(a2 in 1:n.al) for(a1 in 1:n.al)

    for(a3 in 1:n.al) for(a4 in 1:n.al)

    {

        if (a1==a3 & a2==a4)

        {

            P.2[a1,a2,a3,a4] <- f[a1]*f[a2]

            P.1[a1,a2,a3,a4] <- f[a1]*f[a2]*(f[a1]+f[a2])/2

        }

        if (a1==a3 & a2!=a4)

        {

            P.1[a1,a2,a3,a4] <- f[a1]*f[a2]*f[a4]/2

        }

        if (a1!=a3 & a2==a4)

        {

            P.1[a1,a2,a3,a4] <- f[a1]*f[a3]*f[a2]/2

        }

    }

#####julia code

f=Array(Float64,1,2)

f[:,:]=0.5

p1=Array(Float64,nal,nal,nal,nal)

p2=Array(Float64,nal,nal,nal,nal)

p1[:,:,:,:]=0.0

p2[:,:,:,:]=0.0

    for (a2 = 1:nal)  for (a1 = 1:nal)

for (a3 = 1:nal)  for (a4 = 1:nal)

  if (a1==a3 & a2==a4)

     p2[a1,a2,a3,a4] = f[a1]*f[a2]

 p1[a1,a2,a3,a4] = f[a1]*f[a2]*(f[a1]+f[a2])/2.0

  end

  

  if (a1==a3 & a2!=a4)

     p1[a1,a2,a3,a4] = f[a1]*f[a2]*f[a4]/2.0

  end

  

  if (a1!=a3 & a2==a4)

     p1[a1,a2,a3,a4] = f[a1]*f[a3]*f[a2]/2.0

  end

end end

end end 

[Tony Kelman]

The reason is & and | operators are bitwise in Julia, and have a higher precedence than == and !=.

julia> 1 == 1 & 3 == 3

false

julia> 1 == 1 && 3 == 3

true

julia> (1 == 1) & (3 == 3)

true

julia> 1 & 3

1

There are a few issues on this, combining the scalar bitwise and array-elementwise meanings into the same & and | operators is not ideal and can lead to confusion here. When in doubt use extra parentheses around conditionals, and use && or || for scalar comparisons.

[meib...@163.com]

Perfect. Thank you for your help.

在 2015年7月29日星期三 UTC-5上午12:30:59，Tony Kelman写道：