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Bitwise AND

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Bitwise AND

 
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  #1
Sep 11th, 2004
This is giving me a headache...can someone please tell me why these statements ARE NOT equivalent?

Statement 1.)

if(running_result && shell[8 * (x - 1) + (y - 1)].input_val)
running_result = true;
else
running_result = false;


Statement 2.)

running_result &= shell[8 * (x - 1) + (y - 1)].input_val;

-------------------------
The first statement works; the second does NOT. The second always evaluates to false. Why? Both of these vars are bools.
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Re: Bitwise AND

 
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  #2
Sep 13th, 2004
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Originally Posted by iamhe
This is giving me a headache...can someone please tell me why these statements ARE NOT equivalent?
-------------------------
The first statement works; the second does NOT. The second always evaluates to false. Why? Both of these vars are bools.
Maybe the problem is elsewhere.
C++ Syntax (Toggle Plain Text) 
  1. #include <iostream>
  2.  
  3.   int main()
  4.   {
  5.      bool a, b;
  6.  
  7.      //Statement 1.)
  8.      a = false; b = false; std::cout << "a = " << a << ", b = " << b;
  9.      if ( a && b )
  10.   	  a = true;
  11.      else
  12.   	  a = false;
  13.      std::cout << ": a = " << a << std::endl;
  14.  
  15.      a = false; b = true; std::cout << "a = " << a << ", b = " << b;
  16.      if ( a && b )
  17.   	  a = true;
  18.      else
  19.   	  a = false;
  20.      std::cout << ": a = " << a << std::endl;
  21.  
  22.      a = true; b = false; std::cout << "a = " << a << ", b = " << b;
  23.      if ( a && b )
  24.   	  a = true;
  25.      else
  26.   	  a = false;
  27.      std::cout << ": a = " << a << std::endl;
  28.  
  29.      a = true; b = true; std::cout << "a = " << a << ", b = " << b;
  30.      if ( a && b )
  31.   	  a = true;
  32.      else
  33.   	  a = false;
  34.      std::cout << ": a = " << a << std::endl;
  35.      std::cout << std::endl;
  36.  
  37.      //Statement 2.)
  38.      a = false; b = false; std::cout << "a = " << a << ", b = " << b;
  39.      a &= b;
  40.      std::cout << ": a = " << a << std::endl;
  41.  
  42.      a = false; b = true; std::cout << "a = " << a << ", b = " << b;
  43.      a &= b;
  44.      std::cout << ": a = " << a << std::endl;
  45.  
  46.      a = true; b = false; std::cout << "a = " << a << ", b = " << b;
  47.      a &= b;
  48.      std::cout << ": a = " << a << std::endl;
  49.  
  50.      a = true; b = true; std::cout << "a = " << a << ", b = " << b;
  51.      a &= b;
  52.      std::cout << ": a = " << a << std::endl;
  53.      std::cout << std::endl;
  54.  
  55.      return 0;
  56.   }
  57.  
  58.   /* my output
  59.   a = 0, b = 0: a = 0
  60.   a = 0, b = 1: a = 0
  61.   a = 1, b = 0: a = 0
  62.   a = 1, b = 1: a = 1
  63.   
  64.   a = 0, b = 0: a = 0
  65.   a = 0, b = 1: a = 0
  66.   a = 1, b = 0: a = 0
  67.   a = 1, b = 1: a = 1
  68.   */
"One of the methods used by statists to destroy capitalism consists in establishing controls that tie a given industry hand and foot, making it unable to solve its problems, then declaring that freedom has failed and stronger controls are necessary." --Ayn Rand
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Re: Bitwise AND

 
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  #3
Sep 13th, 2004
the first statements uses logical and (&&) the second uses bitwise and (&), so you might try using '&&=' in the second statement.

The difference might be subtle; say the input_val was '2'. In the first statement the running result would be true and in the second statement it would be 0 because TRUE (1) & 2 == 0, but (1 && 2) is true.
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Re: Bitwise AND

 
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  #4
Sep 13th, 2004
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Originally Posted by Chainsaw
the first statements uses logical and (&&) the second uses bitwise and (&), so you might try using '&&=' in the second statement.
There is no such critter: '&&='.
"One of the methods used by statists to destroy capitalism consists in establishing controls that tie a given industry hand and foot, making it unable to solve its problems, then declaring that freedom has failed and stronger controls are necessary." --Ayn Rand
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Re: Bitwise AND

 
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  #5
Sep 13th, 2004
Hee hee - well there SHOULD be! In any case, & and && are not the same.
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Re: Bitwise AND

 
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  #6
Sep 13th, 2004
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Originally Posted by Chainsaw
In any case, & and && are not the same.
But for bools, the results should be the same, shoudn't they?

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Originally Posted by iamhe
Why? Both of these vars are bools.
"One of the methods used by statists to destroy capitalism consists in establishing controls that tie a given industry hand and foot, making it unable to solve its problems, then declaring that freedom has failed and stronger controls are necessary." --Ayn Rand
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Re: Bitwise AND

 
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  #7
Sep 14th, 2004
I am so excited because I finished my program.

Chainsaw is right, any nonzero value is true in a bool, so a bitwise AND wouldn't necessarily evaluate to true... and I shouldn't have tried using a bitwise op for boolean algebra.

(Incidentally I settled on):

running_result = running_result && shell[8 * (x - 1) + (y - 1)].input_val;

Not too stylish, but it served its purpose.
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