If condition A is matched, condition B needs to be matched in order to do action C

Question

My question is:

if (/* condition A */)
{
    if(/* condition B */)
      {
         /* do action C */
      }
    else
      /* ... */
}
else
{
   /* do action C */
}

Is it possible to just write the code of action C one time instead of twice?

How to simplify it?

Answer 1

Your first step in these kinds of problems is always to make a logic table.

A | B | Result
-------------------
T | T | do action C
T | F | ...
F | T | do action C
F | F | do action C

Once you've made the table, the solution is clear.

if (A && !B) {
  ...
}
else {
  do action C
}

Do note that this logic, while shorter, may be difficult for future programmers to maintain.

Answer 2

You have two options:

1. Write a function that performs "action C".

2. Rearrange your logic so that you do not have so many nested if statements. Ask yourself what conditions cause "action C" to occur. It looks to me like it happens when either "condition B" is true or "condition A" is false. We can write this as "NOT A OR B". Translating this into C code, we get

   if (!A || B) {
       action C
   } else {
       ...
   }
   

To learn more about these kind of expressions, I suggest googling "boolean algebra", "predicate logic", and "predicate calculus". These are deep mathematical topics. You don't need to learn it all, just the basics.

You should also learn about "short circuit evaluation". Because of this, the order of the expressions is important to exactly duplicate your original logic. While B || !A is logically equivalent, using this as the condition will execute "action C" when B is true regardless of the value of A.

Answer 3

You can simplify the statement like this:

if ((A && B) || (!A)) // or simplified to (!A || B) as suggested in comments
{
    do C
}

Otherwise put code for 'C' in a separate function and call it:

DoActionC()
{
    ....
    // code for Action C
}
if (condition A)
{
    if(condition B)
    {
        DoActionC(); // call the function
    }
    else
    ...
}
else
{
   DoActionC(); // call the function
}

Answer 4

In a language with pattern matching, you can express the solution in a way that more directly reflects the truth-table in QuestionC's answer.

match (a,b) with
| (true,false) -> ...
| _ -> action c

If you're not familiar with the syntax, each pattern is represented by a | followed by the values to match with (a,b), and the underscore is used as a wildcard to mean "any other values". Since the only case where we want to do something other than action c is when a is true and b is false, we explicitly state those values as the first pattern (true,false) and then do whatever should be done in that case. In all other cases, we fall through to the "wildcard" pattern and do action c.

Answer 5

The problem statement:

If condition A is matched, condition B needs to be matched in order to do action C

describes implication: A implies B, a logical proposition equivalent to !A || B (as mentioned in other answers):

bool implies(bool p, bool q) { return !p || q; }

if (implies(/* condition A */,
            /* condition B */))
{
    /* do action C */
}

Answer 6

Ugh, this tripped me up too, but as pointed out by Code-Apprentice we're guaranteed to need to do action C or run the nested-else block, thus the code could be simplified to:

if (not condition A or condition B) {
    do action C
} else {
    ...
}

This is how we're hitting the 3 cases:

1. The nested do action C in your question's logic required condition A and condition B to be true -- In this logic, if we reach the 2^nd term in the if-statement then we know that condition A is true thus all we need to evaluate is that condition B is true
2. The nested else-block in your question's logic required condition A to be true and condition B to be false -- The only way that we can reach the else-block in this logic would be if condition A were true and condition B were false
3. The outer else-block in your question's logic required condition A to be false -- In this logic if condition A is false we also do action C

Props to Code-Apprentice for straightening me out here. I'd suggest accepting his answer, since he presented it correctly without editing :/

Answer 7

Even though there are already good answers, I thought that this approach might be even more intuitive to someone who is new to Boolean algebra then to evaluate a truth table.

First thing you want to do is look, under which conditions you want to execute C. This is the case when (a & b). Also when !a. So you have (a & b) | !a.

If you want to minimize you can go on. Just like in "normal" arithmetic's, you can multiply out.

(a & b) | !a = (a | !a) & (b | !a). a | !a is always true, so you can just cross it out, which leaves you with the minimized result: b | !a. In case the order makes a difference, because you want to check b only if !a is true (for example when !a is a nullpointer check and b is an operation on the pointer like @LordFarquaad pointed out in his comment), you might want to switch the two.

The other case (/* ... */) is will be always executed when c is not executed, so we can just put it in the else case.

Also worth mentioning is that it probably makes sense either way to put action c into a method.

Which leaves us with the following code:

if (!A || B)
{
    doActionC()  // execute method which does action C
}
else
{
   /* ... */ // what ever happens here, you might want to put it into a method, too.
}

This way you can also minimize terms with more operands, which quickly gets ugly with truth tables. Another good approach are Karnaugh maps. But I won't go deeper into this now.

Answer 8

In the logic concept, you can solve this problem as follow:

f = a.b + !a
f = ?

As a proven problem, this results in f = !a + b. There are a some ways to prove the problem such as truth table, Karnaugh Map and so on.

So in C based languages you can use as follow:

if(!a || b)
{
   // Do action C
}

P.S.: Karnaugh Map is also used for more complicate series of conditions. It's a method of simplifying Boolean algebra expressions.

Answer 9

To make the code look more like text, use boolean flags. If the logic is especially obscure, add comments.

bool do_action_C;

// Determine whether we need to do action C or just do the "..." action
// If condition A is matched, condition B needs to be matched in order to do action C
if (/* condition A */)
{
    if(/* condition B */)
      do_action_C = true; // have to do action C because blah
    else
      do_action_C = false; // no need to do action C because blarg
}
else
{
  do_action_C = true; // A is false, so obviously have to do action C
}

if (do_action_C)
  {
     DoActionC(); // call the function
  }
else
  {
  ...
  }

Answer 10

if((A && B ) || !A)
{
  //do C
}
else if(!B)
{
  //...
}

Answer 11

I would extract C to a method, and then exit the function as soon as possible in all cases. else clauses with a single thing at the end should almost always be inverted if possible. Here's a step by step example:

Extract C:

if (A) {
   if (B)
      C();
   else
      D();
} else
   C();

Invert first if to get rid of first else:

if (!A) {
   C();
   return;
}

if (B)
   C();
else
   D();

Get rid of second else:

if (!A) {
   C();
   return;
}

if (B) {
   C();
   return;
} 

D();

And then you can notice that the two cases have the same body and can be combined:

if (!A || B) {
   C();
   return;
}

D();

Optional things to improve would be:

• depends on context, but if !A || B is confusing, extract it to one or more variables to explain the intent

• whichever of C() or D() is the non-exceptional case should go last, so if D() is the exception, then invert the if one last time

Answer 12

Using flags can also solve this problem

int flag = 1; 
if ( condition A ) {
    flag = 2;
    if( condition B ) {
        flag = 3;
    }
}
if(flag != 2) { 
    do action C 
}