If else problem

[questionit]

Basic if statement question:

If ( .... )
{
If (....) // If 1 ***
If (....) // If 2 ***
If (....) // If 3 ***
}
else
if ( .... )
{
if (something = something)
{
if (....) // *** This and the next 2 If statements are identical to those
if (....) // in code above... how would you avoid duplicating the same
if (....) // if statements.. here are only 3 but what if there are 10s?
}
}

One solution to avoid duplicating the the same if statements is to use goto statement, but is there any othe rbetter and preferred way of doing it?


Thanks

[JonLT]

Basic if statement question:

If ( .... )
{
If (....) // If 1 ***
If (....) // If 2 ***
If (....) // If 3 ***
}
else
if ( .... )
{
if (something = something)
{
if (....) // *** This and the next 2 If statements are identical to those
if (....) // in code above... how would you avoid duplicating the same
if (....) // if statements.. here are only 3 but what if there are 10s?
}
}

One solution to avoid duplicating the the same if statements is to use goto statement, but is there any othe rbetter and preferred way of doing it?


Thanks

you could use a function:

bool ifFunction(bool state)
{
   if(...)
      if(...)
         if(...)
            return true;
 
   return false;
}
 

and then use it like this

If  ( .... ) 
{
   if(ifFunction(...))
}
else 
   if ( .... )
   {
      if (something = something)
      {
         if(ifFunction(...))
      } 
   }
 

[weaknessforcats]

And you could make that function inline to avoid the overhead of a call.

[JosAH]

you could use a function:

bool ifFunction(bool state)
{
   if(...)
      if(...)
         if(...)
            return true;
 
   return false;
}
 

and then use it like this

If  ( .... ) 
{
   if(ifFunction(...))
}
else 
   if ( .... )
   {
      if (something = something)
      {
         if(ifFunction(...))
      } 
   }
 

How about a little Boolean algebra?

T => (A , B , C)
!T => (D , B , C)

rewriting the terms yields

T => A
!T => D
B
C

this reduces easily to code like this:

if (T) A;
else D;
B;
C;
 

kind regards,

Jos

[weaknessforcats]

How about a little Boolean algebra?

Ah yes. DeMorgan's Theorem and Karnaugh Maps.

Check out Introduction to Boolean Algebra and Logic Design by Gerhard Hoernes (McGraw-Hill 1964).

[JosAH]

Ah yes. DeMorgan's Theorem and Karnaugh Maps.

Check out Introduction to Boolean Algebra and Logic Design by Gerhard Hoernes (McGraw-Hill 1964).

Yes, it's fun isn't it? Have a look at the Quine McCluskey (sp?) algorithm for a
better, less human oriented, less visual approach at boolean logic formula
minimization; those Karnaugh maps are a disaster in that respect.

kind regards,

Jos

[weaknessforcats]

less visual approach at boolean logic formula
minimization; those Karnaugh maps are a disaster in that respect.

Not if you use Gray Code.

[JosAH]

Not if you use Gray Code.

Care to elaborate? because I don't understand how, e.g. a 3 variables Karaugh
map (8 cells) would be organized such that a Gray code can come in handy.
It might be just me because I've been lazy all day in my back garden ;-)

kind regards,

Jos (<-- no noticable brain activity to be found)

[weaknessforcats]

In a Gray code only one bit can change state to advance to the next code in the sequence. There are various forms but here is one:

000
001
011
010
110
111
101
100

By using a Gray code on the toroidal Karanugh map and plotting your true states of your Boolean equation, should any two true states be adjacant then one of the variables is irrelevant. That's because it changed state and did not affect the outcome.

Four adjacent true points mean two irrelevant variables.

And, of course, the open sequares are the inverse.

It's much easier for reduction than using a Vietsch diagram and measuring

[JosAH]

In a Gray code only one bit can change state to advance to the next code in the sequence. There are various forms but here is one:

000
001
011
010
110
111
101
100

By using a Gray code on the toroidal Karanugh map and plotting your true states of your Boolean equation, should any two true states be adjacant then one of the variables is irrelevant. That's because it changed state and did not affect the outcome.

Four adjacent true points mean two irrelevant variables.

And, of course, the open sequares are the inverse.

It's much easier for reduction than using a Vietsch diagram and measuring

Yes I know what Gray codes are; but still, finding 'adjacent' areas is still a very
visual process; the Q-McK algorithm simply produces tables of the terrms of
disjunctive normal forms and produces a reduced table (if possible) etc. etc.
until no further reduction is possible (it tries to find 'resolutions' just like those
Gray code numbers). The Karnaugh maps make me dizzy above three (four?)
variables; the Q-McK algorithm doesn't care about many variables are used.

Doing that by hand is a boring and error prone process.
I agree Karnaugh maps are more fun to work (play?) with for humans using just
a few variables.

kind regards,

Jos

[JamC]

Getting back to the question
You can use compound expression in if

Ex..
 
if (age >= 0) 
   if (age < 120)
      print age is valid
 
to
 
if (age >= 0) && (age < 120)
  print age is valid
=========================
 
if (age < 0) || (age >= 120)
  print age is valid
 
to
 
if (age < 0)
   print age is valid
else if (age >= 120)
   peint age is valid