Invert Value or however you want call it...

[Dr. Dri]

If you want to write really obfuscated code you can use this:

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a-a-a

is that normal ?
if i do the same with a C compiler (a -= a - a) a doesn't change.
it's like a -= 0

Dri

[Deeem2031]

If you want to write really obfuscated code you can use this:

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a-a-a

is that normal ?
if i do the same with a C compiler (a -= a - a) a doesn't change.
it's like a -= 0

Dri

Its normal, because the PBCompiler replace "a-" with "a=a-" -> "a=a-a-a" and this is like "a=-a"

[Froggerprogger]

Hmm. In C:

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a -= a-a

is interpreted as subtracting the 'whole' right side of the expression.

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a = a - (a-a)

so

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a = a - a + a = a

in opposite to PB:

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a-a-a

is simply interpretated as

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a = a-a-a = 0-a = -a

So there's really a sematic difference between:

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a -= a-a (C)

and

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a-a-a (PB)

Never thought about, but good to know.
In PB you'll get the C-behaviour by

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a-(a-a)

[Dr. Dri]

if you take a look at the help file, you'll see
variable - expression ;...

so if expression is a - a it shouldn't behave this way

Dri

[Froggerprogger]

Hey, you're right!
The helpfile is inconsistent to the implementation there.
It says (in translation from German):

variable - expression:
The value of the expression is subtracted from variable

That's not true!

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a = 10
debug a-1-1

gives 8, not 10, though the value of the 'expression' is 1-1 = 0

It would be nice to update the docs there, but not change the behaviour.
I find it intuitive and would expect, that the expression a-1-1 gives 8 here, as well I would expect a -= 1-1 to result in 10.

[Trond]

Hey, you're right!
The helpfile is inconsistent to the implementation there.
It says (in translation from German):

variable - expression:
The value of the expression is subtracted from variable

That's not true!

Code: Select all

a = 10
debug a-1-1

gives 8, not 10, though the value of the 'expression' is 1-1 = 0

It would be nice to update the docs there, but not change the behaviour.
I find it intuitive and would expect, that the expression a-1-1 gives 8 here, as well I would expect a -= 1-1 to result in 10.

Everyone seems a bit confused. I can assure that everything is normal and working fine. (Well, actually the manual is a small bit inconsistent with itself.)

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a-a

a-a = 0. Since there is not assignment operator, the result is put into the first of the variable on the left side. The result is that a equals 0.

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a-a-a

Let's say a is 5. That gives us this expression: 5-5-5. Obviously, the result should be -5. Since there is no assignment operator, the result is put into the first of the variables on the left side. A therefore equals -5 after execution.

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a-1-1

Let's say a is 10. That gives the expression 10-1-1 (=8). Since there is no assignment operator the result is put into a.

Then for the "strange" manual. It is not exactly wrong, that depends on how you expect the compiler to handle expressions. You say that 1-1 is 0. That's one way of looking at it. On the other hand, we can say that it is implicit that the expression on the RHS is negative since it follows a minus (and ever number following a single minus is negative) and since the first minus and the second minus has the same precedence level and gets calculated from left to right it is not wrong to assume that the actual expression is -1-1.

But if it was me who wrote the manual I would have rephrased it because as it is now it is more natural to interpret it different from how the compiler interprets it. (Pun unintended.)

[Froggerprogger]

@Trond
I too think, that a-a-a has the correct and expected behaviour, though it is different from a -= a-a

But the manual is 'exactly wrong' in this point. Or at least usually an information in a programming-language-doc given in such a formal way as

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variable - expression

has to be interpretated in a strict formal syntactic way.
It's like a definition made in BNF:
variable and expression are non-terminals, while '-' is a terminal symbol.
so in a-a-a there the is:
variable == a,
'-' == '-'
expression == a-a

So the value of the expression is 0.

For sure this is not really worth the lines I write on this topic, but it is one of the small nice inconsistencies where somebody can be happy to find them and talk about.

[Trond]

It's not that I don't see that it's clearly an error in the manual. It's just that if I say that something minor like that is wrong Dare2 pops up and calls me "Role model for Anal Retentives" so that's why I was a bit modest.

[Dr. Dri]

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a-a-a

Let's say a is 5. That gives us this expression: 5-5-5. Obviously, the result should be -5. Since there is no assignment operator, the result is put into the first of the variables on the left side. A therefore equals -5 after execution.

you point the problem... if "-" is the first operator, then it's also an assignment operator. this is the same behavior with others operators (- * / % & |...)

Dri

[Psychophanta]

asm output code (/commented)

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; a = -a
  MOV    ebx,dword [v_a]
  NEG    ebx
  MOV    dword [v_a],ebx

working "inlined" asm code

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a = 1
!NEG [v_a]
Debug a

I think "NEG a" is faster than using ebx but maybe i'm wrong... It may be optimized as "a + 1" becomes "INC a".

Dri

True, it can be easily optimized, i put it on the todo list

Heeeyy...
Don't for get this too:

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a*-1

and

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a/-1

That should be converted to !NEG [v_a] too
And of course !FCHS if it is a float

[blueznl]

i like role model pictrues especially the... euh... ones...

[Bonne_den_kule]

So much discussion about easy and simple math...

[Psychophanta]

asm output code (/commented)

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; a = -a
  MOV    ebx,dword [v_a]
  NEG    ebx
  MOV    dword [v_a],ebx

working "inlined" asm code

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a = 1
!NEG [v_a]
Debug a

I think "NEG a" is faster than using ebx but maybe i'm wrong... It may be optimized as "a + 1" becomes "INC a".

Dri

True, it can be easily optimized, i put it on the todo list

Still not done in PB4.0

[va!n]

i would like to see fred´s actual todo list