1

I'm learning C++ with previous Java experience.I found the following example in the cplusplus.com :

int CDummy::isitme (CDummy& param)
{
  if (&param == this)
  { 
       return true; //ampersand sign on left side??
  }
  else 
  {    
       return false;
  }
}

My question is : Why does it return boolean false when I clearly declared the method as int ? It's never possible in Java. The link of the example is : here.

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  • 5
    @KeithThompson - The author is writing crap code IMHO Commented Jul 19, 2016 at 23:00
  • 6
    I wouldn't trust an example that takes so long to say return &param == this;. Commented Jul 19, 2016 at 23:00
  • 1
    Perhaps historical reasons? Very early versions of C++, and versions of C prior to C99, didn't have a bool type, so int was commonly used to represent boolean values (0 for false, anything else (especially 1) for true). That's no excuse for writing code like that for a C++ reference site. Commented Jul 19, 2016 at 23:03
  • 1
    @SittingBull I can feel with you. There are much better and reliable resources available than cplusplus.com actually. Commented Jul 19, 2016 at 23:12
  • 1
    @SittingBull You might be interested in reading this post. Commented Jul 19, 2016 at 23:16

2 Answers 2

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8

While the question of why the function does what it does is best answered by the author of the function, it is easy to explain why C++ allows such function to compile without a problem.

In C++ bool is a fundamental integral type, so it can be freely converted to a number: true becomes 1, and false becomes zero.

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Comments

0

This is just an instance of implicit conversion. In c++, 1 and true (also 0 and false) are the same thing(s).

You could also do something like

while (1)
   //Infinite loop

or

bool b = false;
if (b == 0) {
    //This is reached
}
else {
    //This is not
}
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2 Comments

(Java is a bit less messy here, but the second example would afaik also work.)
It is even the case that all numbers other than 0 are truthy

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