[solved] RVAL eval order

[jihem Participant]

This code display -5, then 5.
It seems that RVAL

Monkey

-myStack.Pop()+myStack.Pop()

1	-myStack.Pop()+myStack.Pop()

is evaluated from rigth to left => -10+5 rather than -5+10.
The first Pop done is the one on rigth (rather than the one on left).
Is this a choice? With others languages (Lua,C,C#, Java…) the call order is reversed.

 

 

Monkey

#Import "<std>" Using std.. Class Stack Field s:List<Double> Method New() s=New List<Double> End Method Push(d:Double) Self.s.Add(d) End Method Pop:Double() Local r:Double=0 If Self.s.Count()>0 Then r=Self.s.Last Self.s.RemoveLast() End Return r End End Function Main() Local myStack:=New Stack() myStack.Push(10) myStack.Push(5) Print(-myStack.Pop()+myStack.Pop()) myStack.Push(10) myStack.Push(5) Local s:Double=myStack.Pop(),n:Double=myStack.Pop() Print(n-s) End

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

#Import "<std>" Using std..   Class Stack   Field s:List<Double>     Method New()     s=New List<Double>   End     Method Push(d:Double)     Self.s.Add(d)   End     Method Pop:Double()     Local r:Double=0     If Self.s.Count()>0 Then       r=Self.s.Last       Self.s.RemoveLast()     End         Return r   End End   Function Main()   Local myStack:=New Stack()   myStack.Push(10)   myStack.Push(5)      Print(-myStack.Pop()+myStack.Pop())      myStack.Push(10)   myStack.Push(5)   Local s:Double=myStack.Pop(),n:Double=myStack.Pop()   Print(n-s) End

[Jesse Participant]

I don’t see the problem
this is what you are doing on the first one:
>Print(-5+10)
and the second:
>Print (10-5)
and I am getting
5
5
which is correct.

[jihem Participant]

I am getting:

Mx2cc version 1.1.02

-5
5

Finished debugging app.

which is not correct… if RVAL is evaluated from left to right :-/

I agree with you it should be :
5
5

[Jesse Participant]

That’s weird. I am getting 5 and 5 and I am using 1.1.0.3 which I dowloaded and compiled it from Github from develop section. The about monkey2 in the help says 1.1.02 but mx2cc says 1.1.03.
Thats on OSX.

[jihem Participant]

I’m on Windows 10 and I’m using the monthly build from the patreon download link.
I’ll download the last version from github and try again.
Thanks for the reply

[jihem Participant]

Same result with Mx2xx 1.1.03…

Mx2cc version 1.1.03
-5
5
Finished debugging app.

Have you tried my test ?

Monkey

Function Main() Local myStack:=New Stack() myStack.Push(10) myStack.Push(5) Print(-myStack.Pop()+myStack.Pop()) myStack.Push(10) myStack.Push(5) Local s:Double=myStack.Pop(),n:Double=myStack.Pop() Print(n-s) End

1 2 3 4 5 6 7 8 9 10 11 12

Function Main()   Local myStack:=New Stack()   myStack.Push(10)   myStack.Push(5)      Print(-myStack.Pop()+myStack.Pop())      myStack.Push(10)   myStack.Push(5)   Local s:Double=myStack.Pop(),n:Double=myStack.Pop()   Print(n-s) End

[Jesse Participant]

Are you sure the last example works as is. I get an illegal use of Generic.. when I try to run it.
I had to declare the stack as <int> for it to run.

and I do get the correct results:

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[Jesse Participant]

Ups! I thought you had put the whole code!
But yea, I get the same results.

[TomToad Participant]

The line compiles to

bb_print(bbString((-f0.l_myStack->m_Pop()+f0.l_myStack->m_Pop())));

So it would appear that MinGW is calling the methods in reverse.

This can be a problem as it creates an inconsistency across platforms.  I have confirmed that Android build processes left to right.  Example below prints 5 on Android and -5 on Windows. (Sorry, haven’t figured out yet how to print values on Android without mojo yet)

Monkey

#Import "<std>" #Import "<mojo>" Using std.. Using mojo.. Class Test Extends Window Field gi:int Method OnRender( display:Canvas ) Override gi = 0 App.RequestRender() display.Clear(Color.Black) display.DrawText(-Fun1()+Fun1(),10,10) End Method Fun1:Int() gi = gi + 5 Return gi End Method End Function Main() New AppInstance New Test App.Run() End

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

#Import "<std>" #Import "<mojo>"   Using std.. Using mojo..         Class Test Extends Window Field gi:int Method OnRender( display:Canvas ) Override gi = 0 App.RequestRender() display.Clear(Color.Black) display.DrawText(-Fun1()+Fun1(),10,10) End Method Fun1:Int() gi = gi + 5 Return gi End Method   End   Function Main() New AppInstance New Test App.Run() End

[jihem Participant]

Arg… This seems to be bind to the windows version. I have the same result on another windows PC.
Maybe a bug ???
?

@tomtoad> Yes, you’re right. I understand this is a MingW issue. But this can’t stay as is…
I found documentation about this problem :
http://stackoverflow.com/questions/949433/why-are-these-constructs-using-undefined-behavior

— / —
C99 standard are 6.5 Expressions, §2

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

and 6.5.16 Assignment operators, §4:

The order of evaluation of the operands is unspecified. If an attempt is made to modify the result of an assignment operator or to access it after the next sequence point, the behavior is undefined.

 

Question :
Should Monkey2 leave us with “undefined behavior” (or manage the call order to have the same result on every OS) ?

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[Mark Sibly Keymaster]

–myStack.Pop()+myStack.Pop()

There is no ‘correct’ order for evaluation here – c++ is free to evaluate args of operators in any order (except for ‘And’, ‘Or’ and ‘?’) so mx2 does this too. Ditto with monkey, perhaps even bmx?

Also be very careful with, eg: Blah( ReadInt(),ReadInt(),ReadInt() ) etc! This is generally the ‘classic’ example of this…

[jihem Participant]

Thanks for your help. I will take care to write :

Monkey

Local s:Double=myStack.Pop(),n:Double=myStack.Pop() Print(n-s)

1 2	Local s:Double=myStack.Pop(),n:Double=myStack.Pop()   Print(n-s)

instead of :

Monkey

Print(-myStack.Pop()+myStack.Pop())

1	Print(-myStack.Pop()+myStack.Pop())

This is the first time I got this issue. The others languages I used manage the call from left to right.
I have learnt a lot today ?

[Mark Sibly Keymaster]

Yeah, it’s probably odd if you’ve never encountered it before, but it allows the compiler to make more aggressive optimizations.

For example, there’s an algorithm that can optimally (ie: ‘perfectly’) allocate registers for expression trees simply by going down the ‘busiest’ branch of the tree first, allocating registers as necessary and freeing them after use (I think that’s how it worked…). Blitz3d used something like this from memory.

[Simon Armstrong Participant]

Jihem, in C++ your solution would not help and IMHO would be prone to even more varied compilation results with Debug and Release likely to behave differently.

If it works for monkey2 (comma separated stuff ends up semicolon separated in C++) then it should be documented if it is not already.

[jihem Participant]

Hi, Simon

I have read this : http://en.cppreference.com/w/cpp/language/eval_order
If I understood correctly (I’m not a native english speaker), I ‘m right :

10) In list-initialization, every value computation and side effect of a given initializer clause is sequenced before every value computation and side effect associated with any initializer clause that follows it in the brace-enclosed comma-separated list of initalizers.

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