How to simplify a nested if statement tree

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I have a nested if statement tree that runs a function 16 times (I know), each time sending in a different set of elements to the function.

The function simply returns true or false.

if(!checkSpecial(1,1,1,1)) {
    if(!checkSpecial(1,1,1,0)) {
        if(!checkSpecial(1,1,0,1)) {
            if(!checkSpecial(1,0,1,1)) {
                if(!checkSpecial(0,1,1,1)) {                                
                    if(!checkSpecial(1,1,0,0)) {                            
                        if(!checkSpecial(1,0,0,1)) {                            
                            if(!checkSpecial(0,0,1,1)) {                            
                                if(!checkSpecial(1,0,1,0)) {                            
                                    if(!checkSpecial(0,1,0,1)) {                            
                                        if(!checkSpecial(0,1,1,0)) {                    
                                            if(!checkSpecial(1,0,0,0)) {                    
                                                if(!checkSpecial(0,1,0,0)) {                    
                                                    if(!checkSpecial(0,0,1,0)) {                    
                                                        if(!checkSpecial(0,0,0,1)) {                    
                                                            if(!checkSpecial(0,0,0,0)) {
                                                            }
                                                        }
                                                    }
                                                }
                                            }
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }   
} else {
    // do other stuff
}

as you can see, if the function returns false in every single one of those instances, I want to do other things.

I don't want to do anything if the function returns true.

My question is, I know there has to be a better way of doing this, I'm assuming through some sort of loop, but I'm not aware of what this type of loop would be called or how it would work.

My fix so far:

for (var i = 0; i < 16; i++) { 
   // HELP!
}

Any pointers would be appreciated. thank you.

You can create an array of the binary representation of numbers 0-15, and then check if every one of them returns false when calling the function with the array items as arguments:

const combinations = Array.from(
  { length: 16 },
  (_, i) => (i >>> 0).toString(2).padStart(4, '0').split('').map(Number)
);

console.log(combinations)
/*
if (combinations.every(combination => checkSpecial(...combination) === false)) {
  // every result was false
} else {
  // at least one result wasn't false
}
*/

How would you refactor nested IF Statements?, It is really hard to tell without knowing how the different checks interact. Rigorous refactoring might be in order. Creating a topology of objects that execute the� How to simplify Nested IF statements IF Function. To perform a logical test, the IF function is usually the go-to method. However, due to its parameter IFS and SWITCH. IFS and SWITCH functions have been released in Excel 2016 with Office 365. Each function addresses CHOOSE and VLOOKUP. The

My only suggestion is as follows:

  1. Initialize an array with every permutation

var permutations = [
	[0, 1, 1, 0],
  [1, 1, 1, 0]
];

Excel Nested IF statements, The tutorial explains how to use the nested IF function in Excel to check multiple conditions. It also shows a few good alternatives to using a� Sometimes I'm writing ugly if-else statements in C# 3.5; I'm aware of some different approaches to simplifying that with table-driven development, class hierarchy, anonimous methods and some more. The problem is that alternatives are still less wide-spread than writing traditional ugly if-else statements because there is no convention for that.

Here is an idea with the for loop. Plug this into your VS

     int a = 0;
     int b = 0;
     int c = 0;
     int d = 0;


     for (a = 0; a <= 1; a++)
     {
        for (b = 0; b <= 1; b++)
        {
           for (c = 0; c <= 1; c++)
           {
              for (d = 0; d <= 1; d++)
              {
                 Console.WriteLine($"function({a}, {b}, {c}, {d})");
              }
           }
        }
     }
     Console.ReadKey(); 

How to simplify Nested IF statements, Nested IF formulas can get messy when dealing with too many conditions. In this guide we're going to see how to simplify Nested IF statements. IFS statement as alternative to nested IF function In Excel 2016 and later versions, Microsoft introduced a special function to evaluate multiple conditions - the IFS function. An IFS formula can handle up to 127 logical_test / value_if_true pairs, and the first logical test that evaluates to TRUE "wins":

function* bin(){
  yield 0;
  yield 1;
}
function* word(size,prefix){
  if(!prefix){
    prefix=[];
  }
  if(size){
    var b=bin(),
        n;
    while((n=b.next()) && !n.done){
      prefix[size-1]=n.value;
      yield* word(size-1,prefix);
    }
  } else {
    yield prefix;
  }
}

var w=word(4),
    cond=true,
    n;
while((n=w.next())&&!n.done){
  if(checkSpecial.apply(n.value)){
    cond=false;
    break;
  }
}

if(cond){

} else {

}

Nested 'if' statements can be combined - C# queries, * @description Nested 'if' statements can be simplified by combining them using '&&'. select outer, "These 'if' statements can be combined." It is unnecessary to nest if statements when neither of them has an else part. The code can be written more simply by combining the statements into a single if statement. FII Open Interest Analysis: Tricks to find what FII are buying and selling using open interest - Duration: 40:27. EQSIS Recommended for you. New

Decision Making in C / C++ (if , if..else, Nested if, if , This condition of C else-if is one of the many ways of importing multiple conditions. Decision making statements in programming languages� If the given bpm is outside of our range, we return “none” immediately. This kind of statement is known as a guard clause. We’ve also removed a duplicate statement – before we had two return "none" statements, we now have one. Step 3 - Negate All The Things. I think we can simplify this code by using negation.

Python IfElse, Examples might be simplified to improve reading and basic understanding. Tutorials, references, and examples are constantly reviewed to avoid errors, but we� Nested IF-THEN-ELSE-END IF . The THEN part and the ELSE part, if any, can contain one or more IF-THEN-ELSE-END IF statement in one of the three forms. That is, when you feel it is necessary, you can use as many IF-THEN-ELSE-END IF statements in the THEN part and the ELSE part as you want.

Python Nested If, You can have if statements inside if statements, this is called nested if statements. Example. x = 41 if x > 10: print� In column F & G, we have conditions- Score and Ratings. In column C, we want to enter nested IF formula that will return Ratings. In cell C2, enter formula =IF (B2=1,$G$3,IF (B2=2,$G$4,IF (B2=3,$G$5,IF (B2=4,$G$6)))) Copying down the formula below, we will get the output as below.

Comments
  • 1. Convert i to an 4 bits, stored in an array as 1 and 0. 2. use Function.prototype.apply
  • Without seeing what the function does it's pretty hard to offer advice.
  • I don't see how knowing what the function does matters. It requests 4 inputs of either 1 or zero and then gives a true or false answer. For all it matters, the function can do nothing at all except always return false... the point is how can I simplify the if statements.
  • As @Pointy has said, without knowing what the function achieves it's hard to offer anything solid. The only advice I can offer is to store each permutation in an array and iterate over that instead of writing every if statement.
  • Can you abort once one call returns true?
  • This duplicates some values (eg, [1,0,0,0])