OpenEuphoria

2. Re: Do I need to refactor this code?

• Posted by DerekParnell (admin) Jul 02, 2013
• 1467 views

if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (x = 9) return 1 end if 
if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (y = 10) return 2 end if 
if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (z = 11) return 3 end if 

if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) then 
  if (x = 9) return 1 end if 
  if (y = 10) return 2 end if 
  if (z = 11) return 3 end if 
end if 

Generated Euphoria code is not this sophisticated yet.

You might want to try this approach, depending on how you automate the code generation ...

sequence temp 
temp = {a,b,c,d,e,f,g,h,i,x} 
 
switch temp do 
    case {0,1,2,3,4,5,6,7,8,9} then return 1 end case 
    case {0,1,2,3,4,5,6,7,8,10} then return 2 end case 
    case {0,1,2,3,4,5,6,7,8,11} then return 3 end case 
end switch 

4. Re: Do I need to refactor this code?

• Posted by TheresNoTime Jul 03, 2013
• 1430 views

Nice idea, but it is impossible. Comparison operation can be varied, not only check for equality. I regret that misled. I would like to draw attention to the structure of the code.

6. Re: Do I need to refactor this code?

• Posted by useless_ Jul 03, 2013
• 1378 views

sequence temp 
temp = {a,b,c,d,e,f,g,h,i,x} 
 
switch temp do 
    case {0,1,2,3,4,5,6,7,8,9} then return 1 end case 
    case {0,1,2,3,4,5,6,7,8,10} then return 2 end case 
    case {0,1,2,3,4,5,6,7,8,11} then return 3 end case 
end switch 

Pretty in many ways. Is there a "don't care" value, or a helpful hint to catch a error or fall-thru in the Eu source code, such that

sequence temp 
temp = {0,1,2,3,4,5,6,7,"<12",9} 
 
switch temp do 
    -- "" = type sequence flag = "don't care" (yeas, aka an error) 
    case {0,1,2,3,4,5,6,7,"",9} then return {1,temp} end case 
end switch 

So while Eu didn't process the "<12", temp is otherwise tested and returned to the user in a sequence for extra processing?

Or praps this:

type P_1(integer x) 
    return x >= 0 and x <= 23 
end type 
sequence temp 
temp = {atom,atom,atom,atom,atom,atom,atom,atom,P_1,atom} 
 
switch temp do 
    case {0,1,2,3,4,5,6,7,12,9} then return 1 end case 
end switch 

Or praps there is some way to push a list of types into the case comparisons internally on a case-by-case basis?

I have also had a situation where this would have been extremely valuable in optical pattern recognition or extracting details about motion. It would also be handy in resolving tags in content addressable data, or misspelling, or ?

useless

8. Re: Do I need to refactor this code?

• Posted by TheresNoTime Jul 03, 2013
• 1384 views

Thanks for ideas! Your answers led me to this thought. We read all the lines and find the most frequent combination of two conjunctions (x and y and z). Replace it with a temporary Boolean variable. Declare it before the first combination (x and y and z). Repeat until such combinations will not stay!

Original code:

if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (x = 9) return 1 end if  
if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (y = 10) return 2 end if  
if (a = 0) and (b = 1) and (c = 2) and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (z = 11) return 3 end if  

After 1st iteration:

integer tmp1 = (a = 0) and (b = 1) and (c = 2) 
if tmp1 and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (x = 9) return 1 end if  
if tmp1 and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (y = 10) return 2 end if  
if tmp1 and (d = 3) and (e = 4) and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (z = 11) return 3 end if  

After 2nd iteration:

integer tmp1 = (a = 0) and (b = 1) and (c = 2) 
integer tmp2 = tmp1 and (d = 3) and (e = 4) 
if tmp2 and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (x = 9) return 1 end if  
if tmp2 and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (y = 10) return 2 end if  
if tmp2 and (f = 5) and (g = 6) and (h = 7) and (i = 8) and (z = 11) return 3 end if  

After 3rd iteration:

integer tmp1 = (a = 0) and (b = 1) and (c = 2) 
integer tmp2 = tmp1 and (d = 3) and (e = 4) 
integer tmp3 = tmp2 and (f = 5) and (g = 6) 
if tmp3 and (h = 7) and (i = 8) and (x = 9) return 1 end if  
if tmp3 and (h = 7) and (i = 8) and (y = 10) return 2 end if  
if tmp3 and (h = 7) and (i = 8) and (z = 11) return 3 end if  

After 4th iteration:

integer tmp1 = (a = 0) and (b = 1) and (c = 2) 
integer tmp2 = tmp1 and (d = 3) and (e = 4) 
integer tmp3 = tmp2 and (f = 5) and (g = 6) 
integer tmp4 = tmp3 and (h = 7) and (i = 8) 
if tmp4 and (x = 9) return 1 end if  
if tmp4 and (y = 10) return 2 end if  
if tmp4 and (z = 11) return 3 end if  

That's all! The method is not ideal, but it is universal, infallible and simple to implement.