I need such a condition in ts:
([treeLevel = 0] && [globalVar = GP:R > 0]) || [PIDinRootline = {$pages.2018}]
I wanna show block if page has treelevel=0 and the get var R > 0, or if page id = $pages.2018
It looks like the similar code in php:
if(($treeLevel == 0 && $r > 0) || (pid == number))
The all expression in first brackets should be right, or in second.
Is it exist the method to group it like the previous record or I can only use userfunc?
There is no grouping in TS conditions, but if you need this particular condition from your post I think it is not needed because brackets around && are useless in this case.
(p && q) || r
is exactly the same as
p && q || r
Did you tested it?
How do I model a boolean AND-evaluation in Promela non-atomically ?
I want to model the statement while (flag[j] == true && victim == i) where the AND-evaluation should be done non-atomically.
How is this possible?
loop:
if
:: flag[j] != true -> goto done
:: else
fi;
if
:: victim != i -> goto done
:: else
fi;
/* body of 'while' */
goto loop;
done:
/* after 'while' */
Note, although I don't have a Spin environment in front of me, what makes you think the conditional is atomic in the first place? I'd test with something like:
byte i = 0
i++ >= 0 && i-- >= 0
never { assert (1 == i) }
But, either way, the first code above shows how to make it non-atomic if need be.
I get this error on my insertion sort algorithm:
insertionsort.lpr(19,17) Error: Incompatible types: got "Boolean" expected "LongInt"
Here's the line 19 of my code
while j > 0 and A[j]>key do
I have tried googling all over the internet but i couldn't find any syntax errors or anything.
Here's the full code if it helps :
program instert;
uses crt;
const
N = 5;
var
i:integer;
j:integer;
key:integer;
A : Array[1..N] of Integer;
procedure insertionsort;
begin
for i := 2 to N do
begin
key := A[1];
j:= i - 1;
while j > 0 and A[j]>key do
begin
A[j+1] := A[j] ;
j := j-1;
end;
A[j+1] := key ;
end;
end;
begin
A[1]:= 9;
A[2]:= 6;
A[3]:= 7;
A[4]:= 1;
A[5]:= 2;
insertionsort;
end.
I also get the same error on the bubble sort algorithm i did. Here's the error line
bubblesort.lpr(26,14) Error: Incompatible types: got "Boolean" expected "LongInt"
Here's line 26 of my algorithm:
until flag = false or N = 1 ;
Here's the full code:
program bubblesort;
uses crt;
var
flag:boolean;
count:integer;
temp:integer;
N:integer;
A : Array[1..N] of Integer;
procedure bubblesort ;
begin
Repeat
flag:=false;
for count:=1 to (N-1) do
begin
if A[count] > A[count + 1] then
begin
temp := A[count];
A[count] := A[count + 1];
A[count] := temp;
flag := true;
end;
end;
N := N - 1;
until flag = false or N = 1 ;
end;
begin
A[1]:= 9;
A[2]:= 6;
A[3]:= 7;
A[4]:= 1;
A[5]:= 2;
N := 5;
bubblesort;
end.
In Pascal, boolean operators and and or have higher precedence than the comparison operators >, =, etc. So in the expression:
while j > 0 and A[j] > key do
Given that and has higher precedence, Pascal sees this as:
while (j > (0 and A[j])) > key do
0 and A[j] are evaluated as a bitwise and (since the arguments are integers) resulting in an integer. Then the comparison, j > (0 and A[j]) is evaluated as a boolean result, leaving a check of that with > key, which is boolean > longint. You then get the error that a longint is expected instead of the boolean for the arithmetic comparison.
The way to fix it is to parenthesize:
while (j > 0) and (A[j] > key) do ...
The same issue applies with this statement:
until flag = false or N = 1 ;
which yields an error because or is higher precedence than =. So you can parenthesize:
until (flag = false) or (N = 1);
Or, more canonical for booleans:
until not flag or (N = 1); // NOTE: 'not' is higher precedence than 'or'
When in doubt about the precedence of operators, parenthesizing is a good idea, as it removes the doubt about what order operations are going to occur.
I have this code:
count = $content.find('.post').length;
for x in [1...count]
/*
prev_el_height += $("#content .post:nth-child(" + x + ")").height();
*/
prev_el_height += $content.find(".post:nth-child(" + x + ")").height();
I expected this to turn into
for (x = 1; x < count; x++) { prev_el ... }
but it turns into this:
for (x = 1; 1 <= count ? x < count : x > count; 1 <= count ? x++ : x--) {
Can somebody please explain why?
EDIT: How do I get my expected syntax to output?
In CoffeeScript, you need to use the by keyword to specify the step of a loop. In your case:
for x in [1...count] by 1
...
You're asking to loop from 1 to count, but you're assuming that count will always be greater-than-or-equal-to one; the generated code doesn't make that assumption.
So if count is >= 1 then the loop counter is incremented each time:
for (x = 1; x < count; x++) { /* ... */ }
But if count is < 1 then the loop counter is decremented each time:
for (x = 1; x > count; x--) { /* ... */ }
Well, you want x to go from 1 to count. The code is checking whether count is bigger or smaller than 1.
If count is bigger than 1, then it has to increment x while it is smaller than count.
If count is smaller than 1, then it has to decrement x while it is bigger than count.
For future reference:
$('#content .post').each ->
prev_el_height += $(this).height()
Has the same effect, assuming :nth-child is equivalent to .eq(), and x going past the number the elements is a typo.
I've got three boolean values A, B and C. I need to write an IF statement which will execute if and only if no more than one of these values is True. In other words, here is the truth table:
A | B | C | Result
---+---+---+--------
0 | 0 | 0 | 1
0 | 0 | 1 | 1
0 | 1 | 0 | 1
0 | 1 | 1 | 0
1 | 0 | 0 | 1
1 | 0 | 1 | 0
1 | 1 | 0 | 0
1 | 1 | 1 | 0
What is the best way to write this? I know I can enumerate all possibilities, but that seems... too verbose. :P
Added: Just had one idea:
!(A && B) && !(B && C) && !(A && C)
This checks that no two values are set. The suggestion about sums is OK as well. Even more readable maybe...
(A?1:0) + (B?1:0) + (C?1:0) <= 1
P.S. This is for production code, so I'm going more for code readability than performance.
Added 2: Already accepted answer, but for the curious ones - it's C#. :) The question is pretty much language-agnostic though.
how about treating them as integer 1's and 0's, and checking that their sum equals 1?
EDIT:
now that we know that it's c#.net, i think the most readable solution would look somewhat like
public static class Extensions
{
public static int ToInt(this bool b)
{
return b ? 1 : 0;
}
}
the above tucked away in a class library (appcode?) where we don't have to see it, yet can easily access it (ctrl+click in r#, for instance) and then the implementation will simply be:
public bool noMoreThanOne(params bool[] bools)
{
return bools.ToList().Sum(b => b.ToInt()) <= 1;
}
...
bool check = noMoreThanOne(true, true, false, any, amount, of, bools);
You shold familiarize yourself with Karnaugh maps. Concept is most often applied to electronics but is very useful here too. It's very easy (thought Wikipedia explanation does look long -- it's thorough).
(A XOR B XOR C) OR NOT (A OR B OR C)
Edit: As pointed out by Vilx, this isn't right.
If A and B are both 1, and C is 0, A XOR B will be 0, the overall result will be 0.
How about:
NOT (A AND B) AND NOT (A AND C) AND NOT (B AND C)
If you turn the logic around, you want the condition to be false if you have any pair of booleans that are both true:
if (! ((a && b) || (a && c) || (b && c))) { ... }
For something completely different, you can put the booleans in an array and count how many true values there are:
if ((new bool[] { a, b, c }).Where(x => x).Count() <= 1) { ... }
I'd go for maximum maintainability and readability.
static bool ZeroOrOneAreTrue(params bool[] bools)
{
return NumThatAreTrue(bools) <= 1;
}
static int NumThatAreTrue(params bool[] bools)
{
return bools.Where(b => b).Count();
}
There are many answers here, but I have another one!
a ^ b ^ c ^ (a == b && b == c)
A general way of finding a minimal boolean expression for a given truth table is to use a Karnaugh map:
http://babbage.cs.qc.edu/courses/Minimize/
There are several online minimizers on the web. The one here (linked to from the article, it's in German, though) finds the following expression:
(!A && !B) || (!A && !C) || (!B && !C)
If you're going for code readability, though, I would probably go with the idea of "sum<=1". Take care that not all languages guarantee that false==0 and true==1 -- but you're probably aware of this since you've taken care of it in your own solution.
Good ol' logic:
+ = OR
. = AND
R = Abar.Bbar.Cbar + Abar.Bbar.C + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar.(Cbar + C) + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar + Abar.B.Cbar + A.Bbar.Cbar
= Abar.Bbar + CBar(A XOR B)
= NOT(A OR B) OR (NOT C AND (A XOR B))
Take the hint and simplify further if you want.
And yeah, get your self familiar with Karnaugh Maps
Depends whether you want something where it's easy to understand what you're trying to do, or something that's as logically simple as can be. Other people are posting logically simple answers, so here's one where it's more clear what's going on (and what the outcome will be for different inputs):
def only1st(a, b, c):
return a and not b and not c
if only1st(a, b, c) or only1st(b, a, c) or only1st(c, a, b):
print "Yes"
else:
print "No"
I like the addition solution, but here's a hack to do that with bit fields as well.
inline bool OnlyOneBitSet(int x)
{
// removes the leftmost bit, if zero, there was only one set.
return x & (x-1) == 0;
}
// macro for int conversion
#define BOOLASINT(x) ((x)?1:0)
// turn bools a, b, c into the bit field cba
int i = (BOOLASINT(a) << 0) | BOOLASINT(b) << 1 | BOOLASINT(c) << 2;
if (OnlyOneBitSet(i)) { /* tada */ }
Code demonstration of d's solution:
int total=0;
if (A) total++;
if (B) total++;
if (C) total++;
if (total<=1) // iff no more than one is true.
{
// execute
}