Will every && run before if on the same line in Ruby?
For example:
#building.approved = params[:approved] if xyz && abc && mno...
Can an unlimited number of && be used on the right side of an if without using parentheses?
I'm inclined to use parentheses but I'd like to understand the default behaviour.
Everything after the if must be part of the condition by virtue of the syntax. The only way to get around this is to be really specific:
(#building.approved = params[:approved] if xyz) && abc && ...
Which is obviously not what you're intending here.
Operator binding strength isn't an issue here since if is a syntax element not an operator, so it has the absolute lowest priority.
The only conditions that will be evaluated are the ones that produce a logically false value, or come after one that was logically true as the first one to return logically false will halt the chain.
That is:
if a && b && c
Will stop at a if that is a logically-false value. b and c will not be evaluated. There's no intrinsic limit on chaining though.
Related
pseudocode:
// deprecated x!=y && hash(x) == hash(y) // how to make this true?
x!=y && hash(x) == hash(y) && (z!=x && z!=y) && (hash(x) != hash(z) && (hash(y) != hash(z)) // how to make this true?
x and y can be any readable value
Whatever the language, the pseudocode is just help to understand what I mean.
I just wonder how to implement such hash function.
PS: For math, i am an idiot. I can not imagine if there is an algorithm that can do this.
UPDATE 1:
The pseudocode has bug, so I updated the code(actually still has bug, never mind, I will explain).
My original requirement is to make a hash function that can return same value for different parameter, and the parameter value should contains some rule. It means, only the parameter value in same category would gets same hash code, others are not.
e.g.
The following expressions are clearly(you can treat '0' as placeholder):
hash("1.1") == hash("1.0") == hash("0.1")
hash("2.2") == hash("2.0") == hash("0.2")
and
hash("2.2") != hash("2.1") != hash("1.2")
I think this question can do such description:
There are two or more different values contains implied same attribute.
Only these values have such same attribute in the world.
The attribute can obtain through some way(maybe a function), hash() will call it inside.
hash() one of the values, you can retrive the attribute, then you can get the unique hashCode.
It's looks like hash collision, but we exactly know what they are. Also looks like many-to-one model.
How to design collision rules? The values could be any character or numeric. And how to implement the designs?
PPS: This is a question full of bugs, maybe the updated parts cannot explain the the problem either. Or maybe this is a false proposition. I want abstract my issue as a general model, but it makes my mind overflowed. If necessary I will post my actual issue that I am facing.
Any constant hash trivially satisfies your condition:
hash(v) = 42
A less constant answer than yuri kilocheck's would be to use the mod operator:
hash(v) = v % 10;
Then you'll have:
hash(1) = 1
hash(2) = 2
hash(3) = 3
...
hash(11) = 1
hash(12) = 2
I have the following code:
until (#world.exists? decision || decision == '')
UiHandler.print_error(UiHandler::NO_TILE)
UiHandler.print_turn_message
decision = gets.chomp
end
which should allow the player to skip a turn by entering an empty line. But for some reason the until loop keeps running even when the condition is true
i.e. passing in '1 1' does work and stop the loop, since it exists in world, but passing nothing doesn't, even though puts (#world.exists? decision || decision == '') gives 'true'
What would cause an until loop not to stop even when the condition is met?
Fix is
(#world.exists?(decision) || decision == '')
Otherwise - #world.exists? decision || decision == '' is being treated as #world.exists?(decision || decision == ''), which is not correct expression, you intended to write.
As decision is a string object, which in Ruby is considered as truth value, decision || decision == '' (in the code written by you) will be evaluated as true too. This decision will be passed as a method argument to the method #world.exists? always.
I'm working on a project. Currently I have a fairly large conditional statement, that assigns a value to a variable based on some input parameters. So, I have something like this.
if some condition
x = some value
elsif another condition
x = a different value
...
What's the best way to refactor this? I'm hoping that I might end up with something like
x = some value if some condition || another value if another condition
Is there a pattern for this sort of thing?
Just put the assignment outside the if.
x = if some condition
some value
elsif another condition
a different value
Or you could use a Hash.
x = dict[some condition]
It's not a pattern, but an operator. The one you're referring to is the ternary operator:
If Condition is true ? Then value X : Otherwise value Y
Here is an example:
speed = 90
speed > 55 ? puts("I can't drive 55!") : puts("I'm a careful driver")
Using the ternary statement is short, sweet, and does the job.
x = some condition ? some value :
another condition ? a different value : ...
A conditional statement is also an expression, so one of the first things you can do, if the variable is the same in each condition, is:
x = if cond1
expr1
elsif cond2
expr2
....
end
If the conditions are all states of a single expression, you can make this even neater, using a case statement.
However, the next most obvious re-factoring exercise is to get the big conditional isolated into a method, which should be fed the bare minimum data required to evaluate all the conditions and expressions.
E.g.
# Where conditional is currently, and x assigned, assuming the conditionals
# need a couple of variables . . .
x = foo param1, param2
# Elsewhere
private
def foo p1, p2
if cond1
expr1
elsif cond2
expr2
....
end
end
If you want to refactor for code clarity and flexibility, consider the replacing conditional with polymorphism refactor.
There's not enough detail in your question to go much further with recommendations, but this refactor will make your code base much more resistant to change. If you receive a new requirement, it's bad form to break open the conditional and modify it (more prone to introducing bugs, more difficult to do); it's preferable to create a new object that you can plug into the existing codebase. This flexibility what the Open/Closed Principle (the "O" in the SOLID acronym) describes.
When programming in Ruby I quite often have assignments like the following
test = some_function if some_function
With that assignments I want to assign the output of a function, but if it returns nil I want to keep the content of the variable. I know there are conditional assignments, but neither ||= nor &&= can be used here. The shortest way I found to describe the statement above is
test = (some_function or test)
Is there a better / shorter way to do this?
I don't think there's anything better than the last snippet you showed but note that or is used for flow control, use || instead:
test = some_function || test
It's usually better to assign new values to new names, the resulting code is easier to understand and debug since variables/symbols have the same value throughout the scope:
some_different_and_descriptive_name_here = some_function || test
I'd just add parentheses
(a = b) unless b.nil?
(a = b) if b
being inferior because if b is false then a remains as before
Keep in mind that this evaluates b twice, so if b is a function with side-effects (such as changing variables outside of its scope or printing) it will do that twice; to avoid this you must use
temp = b; (a = temp) unless temp.nil?
(which can, of course, be split into)
temp = b
(a = temp) unless temp.nil?
It's a bit confusing to me about what is the difference between these condition expressions below:
if( 1 == a) {
//something
}
and
if( a == 1 ) {
//something
}
I saw the above one in some scripts I have downloaded and I wonder what's the difference between them.
The former has been coined a Yoda Condition.
Using if(constant == variable) instead of if(variable == constant), like if(1 == a). Because it's like saying "if blue is the sky" or "if tall is the man".
The constant == variable syntax is often used to avoid mistyping == as =. It is, of course, often used without understanding also when you have constant == function_call_retuning_nothing_modifiable.
Other than that there's no difference, unless you have some weird operator override.
Many programming languages allow assignments like a = 1 to be used as expressions, making the following code syntactically valid (given that integers can be used in conditionals, such as in C or many scripting languages):
if (a = 1) {
// something
}
This is rarely desired, and can lead to unexpected behavior. If 1 == a is used, then this mistake cannot occur because 1 = a is not valid.
Well, I am not sure about the trick. Generally, we could say the equal sign is commutative. So, a = b implies b = a. However, when you have == or === this doesn't work in certain cases, for example when on the right side you have a range: 5 === (1..10) vs. (1..10) === 5.